05AB1E, 52 48 bytes

Δ9EDNQ©˜ƶ5äΔ2Fø0δ.ø}2Fø€ü3}®*€€à}Иs¢2@*_*}ø0δ†ø 

Uses 0 for smoke.

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Explanation:

Δ # Loop until the result no longer changes: 9E # Loop `N` in the range [1,9]: D # Duplicate the current matrix # (which will be the implicit input-matrix in the first iteration) NQ # Check which values are equal to `N` in the copy © # Store this matrix of 1s/0s in variable `®` (without popping) ˜ # Flatten it to a list ƶ # Multiply each value by its 1-based index 5ä # Convert the list back to a 5x5 matrix Δ # Loop until it no longer changes to flood-fill the positive values: 2Fø0δ.ø} # Add a border of 0s around the matrix: 2F } # Loop 2 times: ø # Zip/transpose; swapping rows/columns δ # Map over each row: 0 .ø # Add a leading/trailing 0 2Fø€ü3} # Convert it into overlapping 3x3 blocks: 2F } # Loop 2 times again: ø # Zip/transpose; swapping rows/columns € # Map over each inner list: ü3 # Convert it to a list of overlapping triplets ®* # Multiply each 3x3 block by the value in matrix `®` # (so the 0s remain 0s) €€à # Keep the largest value from each 3x3 block: €€ # Nested map over each 3x3 block: à # Pop and push its flattened maximum } # Close the flood-fill loop Иs¢2@*_* # Remove all islands of 2 or more cells from the matrix: Ð # Triplicate the resulting matrix of 0s and positive islands ˜ # Flatten the top copy s # Swap so the other copy of the matrix is at the top ¢ # For each value in the matrix, count its occurrences in the list 2@ # Check for each count whether it's >= 2 * # Multiply these checks to the matrix of 0s and positive islands _ # Check for each value whether it's equal to 0 # (the falsey results are the islands we'll remove) * # Multiply it to the values of the matrix to 'explode' the islands } # Close the [1,9]-loop ø0δ†ø # Collapse the bricks of the building down after the explosions: ø # Zip the matrix; swapping rows/columns δ # Map over each inner list (each column): 0 † # Filter all 0s to the front ø # Zip/transpose it back # (after which the result is output implicitly) 

JavaScript (ES7), 167 bytes

Expects a matrix of integers. Returns another matrix of integers, with 0 for smoke.

f=(m,V=[],X,Y)=>~m.map((r,y)=>r.map((v,x)=>X+1?6>>(X-x)**2+(Y-y)**2&V==v?V='':0:V[x]=f(m,v,x,y)+[V[x]]))/X?V:m+''==(V.map((s,j)=>m.map((r,i)=>r[j]=~~s[4-i])),m)?m:f(m) 

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How?

We use the same recursive function to process two distinct steps:

1. Fill an array of strings representing the updated columns of the matrix without removed cells. Then convert this array back to a matrix.
2. Test whether a given cell should be removed.

Step 1

The variable \$V\$ is initially set to an empty array and the variables \$X\$ and \$Y\$ are undefined.

For each cell at \$(x,y)\$ with value \$v\$, we insert the result of the second step at the beginning of \$V[x]\$ (coerced to an empty string if it's still undefined):

V[x] = f(m, v, x, y) + [V[x]] 

Once all cells have been processed, we transpose the strings stored in \$V\$ to build the new matrix with each column filled with leading 0's wherever needed:

V.map((s, j) => m.map((r, i) => r[j] = ~~s[4 - i]) ) 

If the matrix is unchanged, we stop and return it. Otherwise, we do a recursive call at step 1.

Step 2

The variable \$V\$ is set to the value of the cell being tested and the variables \$X\$ and \$Y\$ hold its coordinates.

We walk through each cell at \$(x,y)\$ and set \$V\$ to an empty string as soon as a neighbor cell with the same value \$v\$ is detected. A neighbor cell is a cell whose squared Euclidean distance with the reference cell is either \$1\$ or \$2\$. This is tested with a bitmask:

6 >> (X - x) ** 2 + (Y - y) ** 2 & V == v 

NB: Because the matrix is \$5\times5\$, the maximum squared Euclidean distance is \$32\$, in which case the shift amount will wrap to \$0\$. Fortunately, this will still work as expected.

Once all cells have been processed, we simply return \$V\$.

Charcoal, 55 bytes

Ｅ⁵ＳＷ¬⁼υθ«≔υθ≔Ｅ⁵⟦⟧υＦ⁵Ｆ⁵«Ｊκ⁻⁴λ¿¬№ＫＭＫＫ⊞§υκＫＫ»⎚ＧT⁵☁↑Ｅυ⪫κω 

Try it online! Link is to verbose version of code. Explanation:

Ｅ⁵Ｓ 

Copy the input to the canvas.

Ｗ¬⁼υθ« 

Repeat until two iterations produce the same result.

≔υθ 

Save the result of the previous iteration.

≔Ｅ⁵⟦⟧υ 

Start collecting lonely bricks for this iteration.

Ｆ⁵Ｆ⁵«Ｊκ⁻⁴λ 

Loop over the columns from left to right and the rows from bottom to top.

¿¬№ＫＭＫＫ⊞§υκＫＫ 

If this brick is lonely then collect it.

»⎚ 

Clear the canvas.

ＧT⁵☁ 

Output a background of clouds, leaving the cursor at the bottom left.

↑Ｅυ⪫κω 

Output the lonely bricks from left to right and bottom to top.

Julia 1.8, 144 bytes

!o=(k=keys(o)).|>_->(o.=[(2>count(o[i].==get.([o],i-k[7].+k[1:3,1:3],0)))o[i] for i=k];[o[j,i]<1&&(o[j-1:j,i]=o[[j,j-1],i]) for i=1:5,j=5:-1:2]) 

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mutates the input. uses zeros for ashes

Python3, 507 bytes:

E=enumerate def S(b): q=[*b] while q: Q=[q.pop(0)];R=[Q[0]] while Q: x,y=Q.pop(0) F=1 for X,Y in[(1,0),(0,1),(0,-1),(-1,0),(-1,-1),(1,-1),(-1,1),(1,1)]: C=(x+X,y+Y) if C in b and b[C]==b[(x,y)] and C not in R:Q+=[C];F=0;R+=[C];q.remove(C) if F and len(R)>1:yield R def f(b): while 1: F=1 for i in S({(x,y):v for x,r in E(b)for y,v in E(r)if v}): F=0 for x,y in i:b[x][y]=0 if F:return b b=[*map(list,zip(*[[j for j in i if 0==j]+[*filter(None,i)]for i in zip(*b)]))] 

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Jelly, 27 24 bytes

ŻṖ;"⁺Ḋ)Zð⁺ċ""Ị×ZṠÞ€Zµ$ƬṪ 

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-2 with ¬¬$ -> Ṡ, -1 with dyadic chaining via extreme chain separator abuse

The neighborhood logic feels a bit... unhinged, but I'm not mentally present enough to devise or steal anything better at the moment.

 ) For each row: ŻṖ Prepend 0 and remove the last element, ;" concatenate with corresponding original elements, ⁺ and do the same Ḋ with the first element removed. Z Transpose, ð⁺ and do it again. ċ"" How many times is the middle in each neighborhood? Ị× Zero out elements with counts greater than 1. Z €Z For each column, Þ sort its elements by Ṡ their signs. µ$ƬṪ Repeat all of that while unique. 

Python 3.8, 231 227 bytes

def f(a,X=range(5)): while(b:=[[(i:=a[r][c])-i*any(all([r-2<R<r+2,c-2<C<c+2,(r,c)!=(R,C),a[R][C]==i])for R in X for C in X)for c in X]for r in X])!=a:a=[[([0]*5+[R[c]for R in b if R[c]])[-5:][r]for c in X]for r in X] return a 

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Ungolfed and commented

def f(a): for _ in range(25):a=h(g(a)) return a # Replace connected integers with 0s def g(a): b=[[*i]for i in a] for r in range(5): for c in range(5): i=a[r][c] if I(a,r,c,i):b[r][c]=0 else:b[r][c]=i return b # Check surrounding cells def I(a,r,c,i): for R in range(5): for C in range(5): if r-2<R<r+2and c-2<C<c+2and(r,c)!=(R,C)and a[R][C]==i:return 1 return 0 # Drop numbers down based on positions of 0 def h(a): b=[[0]*5for i in a] for c in range(5): C=([0]*5+[R[c]for R in a if R[c]])[-5:] for r in range(5):b[r][c]=C[r] return b 

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