R, 50 40 bytes

-4 bytes thanks to Giuseppe.

k=scan();while(grepl("[2-9]",F<-F+k))0;F 

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The while statement goes through all multiples of k and searches for any digits greater than 1. The advantage of using grepl (over utf8ToInt or strsplit) is that it accepts integer input and converts it to string.

R, 50 38 bytes

-4 bytes thanks to Giuseppe.

grep("^[01]+$",(k=scan())*1:10^k)[1]*k 

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It follows from this blog post (linked in the question) that the smallest binary multiple of \$k\$ is smaller than \$2\cdot10^{k-1}\$; this answer uses the larger bound \$k\cdot10^k\$ instead.

Creates a vector of all multiples of \$k\$ between \$k\$ and \$k\cdot10^k\$. The regexp gives the indices of those made only of 0s and 1s; select the first index and multiply by \$k\$ to get the answer.

Will time out on TIO for input greater than 8, but with infinite memory it would work for any input.

R, 50 44 bytes

k=scan();while(sum(grep("[2-9]",F<-F+k)))0;F 

Try it online!

The while statement goes through all multiples of k and searches for any digits greater than 1. The sum is necessary as when there is no match, grep outputs an empty vector; sum transforms this to 0. The advantage of using grep (over utf8ToInt or strsplit) is that it accepts integer input and converts it to string.

R, 50 40 bytes

-4 bytes thanks to Giuseppe.

k=scan();while(grepl("[2-9]",F<-F+k))0;F 

Try it online!

The while statement goes through all multiples of k and searches for any digits greater than 1. The advantage of using grepl (over utf8ToInt or strsplit) is that it accepts integer input and converts it to string.

R, 50 bytes

k=scan();while(any(utf8ToInt(c(F<-F+k,""))>49))0;F 

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Recall that the ASCII codepoints of 0 and 1 are 48 and 49.

The while statement goes through all multiples of k, converting each to character with c(…,"") and checking whether any codepoints are greater than 49, i.e. whether there any digits greater than 1.

R, 50 44 bytes

k=scan();while(sum(grep("[2-9]",F<-F+k)))0;F 

Try it online!

The while statement goes through all multiples of k and searches for any digits greater than 1. The sum is necessary as when there is no match, grep outputs an empty vector; sum transforms this to 0. The advantage of using grep (over utf8ToInt or strsplit) is that it accepts integer input and converts it to string.