Distance between powers of 2 and 3 As we know $3^1-2^1 = 1$ and of course $3^2-2^3 = 1$. The question is that whether set $$ \{\ (m,n)\in \mathbb{N}\quad |\quad |3^m-2^n| = 1 \} $$
is finite or infinite.

Note first that if $3^{2n}-1=2^r$ then $(3^n+1)(3^n-1)=2^r$. The two factors in brackets differ by $2$ so one must be an odd multiple of $2$, and this is only possible if $n=1$ (the only odd number we can allow in the factorisation is $1$)
Now suppose that $3^n-1=2^r$ and $n$ is odd. Now $3^n\equiv -1$ mod $4$ so $3^n-1$ is not divisible by $4$.
Now suppose $3^n=2^{2r}-1=(2^r+1)(2^r-1)$. The two factors differ by $2$ and cannot therefore both be divisible by $3$. Only $r=1$ is possible.
The final case is $3^n=2^{k}-1$ where $k$ is odd. Now the right hand side is $\equiv -2$ mod $3$, so only $k=1$ is possible, and $n=0$ (if permitted).
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The previous version of this answer was overcomplicated - trying to do things in a hurry.
