A proof of $\dim(R[T])=\dim(R)+1$ without prime ideals?

Let $R$ be a commutative ring. For $x\in R$ write $R_{\{x\}}$ for the localization at the multiplicative set $x^{\mathbb{N}}(1+xR)\subset R$. As requested, we use the following characterization.

Coquand--Lombardi characterization. For every ring $A$, $$ \operatorname{dim}(A)=\sup_{x\in A}\bigl(\operatorname{dim}(A_{\{x\}})+1\bigr). $$ Equivalently, $\operatorname{dim}(A)\leqslant k$ if and only if for all $x_0,\dots,x_k\in A$ there exist $m_0,\dots,m_k\in\mathbb{N}$ and $a_0,\dots,a_k\in A$ such that, defining recursively $$ \begin{aligned} & E_k(y_k):=y_k^{m_k}\bigl(1+a_k y_k\bigr),\\ & E_j(y_j,\dots,y_k):=y_j^{m_j}\bigl(E_{j+1}(y_{j+1},\dots,y_k)+a_j y_j\bigr)\quad(j=k-1,\dots,0), \end{aligned} $$ one has $$ E_0(x_0,\dots,x_k)=0. $$

Claim. If $R$ is Noetherian, then $\operatorname{dim}(R[T])=\operatorname{dim}(R)+1$.

By the characterization with $A=R[T]$, it suffices to show that $\operatorname{dim}\bigl(R[T]_{\{f\}}\bigr)\leqslant \operatorname{dim}(R)$ for every $f\in R[T]$. The opposite inequality $\operatorname{dim}(R[T])\geqslant \operatorname{dim}(R)+1$ is elementary and holds for all rings.

Step 1 (monicization inside $R[T]_{\{f\}}$, no primes). Let $f(T)=a_dT^d+\cdots+a_0$ with $d\geqslant 1$ and leading coefficient $a_d$. Let $S$ be the submonoid of $R[T]$ generated by $f$ and the elements $(1+fh)$ with $h\in R[T]$, and work in the localization $B:=S^{-1}R[T]=R[T]_{\{f\}}$. If $R$ is Noetherian, the Dedekind--Mertens content formula applied to $F(U):=f(T+U)\in (R[T])[U]$ and $H(U):=(1+U)^d-1$ gives an identity $$ a_d^{N}=u\cdot f+\sum_{i} v_i\bigl((1+f h_i)^d-1\bigr), $$ for some $N\in\mathbb{N}$ and $u,v_i,h_i\in R[T]$. In $B$ the elements $f$ and each $(1+fh_i)$ are units, hence $a_d$ is a unit in $B$. Therefore $f$ is associated in $B[T]$ to a monic polynomial $p$ of the same degree, and $R[T]_{\{f\}}=R[T]\{p\}$. From now on assume $f$ is monic.

Step 2 (a monic divisor does not increase the dimension). Let $A$ be any ring and let $p\in A[T]$ be monic of degree $d\geqslant 1$. Set $C:=A[T]\{p\}$. Take arbitrary $y_0,\dots,y_k\in C$ with $k\geqslant 0$. Clear denominators: there is $s$ in the monoid generated by $p$ and the $(1+ph)$ with $h\in A[T]$ such that $z_i:=s\,y_i\in A[T]$. Divide by $p$ (possible since $p$ is monic): write $z_i=q_i p+r_i$ with $\deg r_i<d$.

Write $A_d:=A[T]/(T^d)$ and view each remainder $r_i$ in $A_d$. Using the CL characterization one checks that $\operatorname{dim}(A_d)=\operatorname{dim}(A)$ (indeed, for $x\in A$ one has $(A_d)_{\{x\}}\cong (A_{\{x\}})[T]/(T^d)$, so the supremum $\sup_x$ is unchanged). If $\operatorname{dim}(A)\leqslant k$, the CL criterion in $A_d$ applied to $r_0,\dots,r_k$ gives exponents $m_i$ and coefficients $a_i$ with $$ E_0(r_0,\dots,r_k)=0\quad\text{in }A_d. $$ Since $\deg E_0(r_0,\dots,r_k)<d$, we have $E_0(r_0,\dots,r_k)=0$ already in $A[T]$. For the same recursive expression built from $z_i=q_ip+r_i$ one has $$ E_0(z_0,\dots,z_k)=p\cdot W(T)\in (p). $$ As $z_i=s\,y_i$ and $s,p$ are units in $C$, it follows that $E_0(y_0,\dots,y_k)=0$ in $C$. This is exactly the CL condition for $\operatorname{dim}(C)\leqslant k$.

Conclusion. Given $f\in R[T]$, Step 1 reduces to the monic case and Step 2 with $A=R$ yields $\operatorname{dim}\bigl(R[T]_{\{f\}}\bigr)\leqslant \operatorname{dim}(R)$. Taking the supremum over $f$ in the Coquand--Lombardi formula for $A=R[T]$ gives $\operatorname{dim}(R[T])\leqslant \operatorname{dim}(R)+1$. The reverse inequality is elementary, hence $\operatorname{dim}(R[T])=\operatorname{dim}(R)+1$.