$2^n-3^m=1 , m,n \in \mathbb N =?$

Question

$2^n-3^m=1 , m,n \in \mathbb N =?$ my questions are:

1. do m,n exist?
2. are they finitely many $m,n$?
3. if there are infinitely many is there a way to describe them all?

Same question about $3^n-2^m=1 $, besides $m=n=1$ are there any others?

PS: This is not from number theory or home work, just personal pondering

Answer 1

[Update: Table & text a bit extended]

I like to look at such questions in a, say, "constructive" way. First I change to the more familiar form $$ 2^n = 3^m + 1$$ Now let m vary, we get, where N is the exponent at primefactor 2 in $3^m+1 (=x \cdot 2^N) $ $$ \begin{array} {r|llll} m & 0&1&2&3&4 & ... \\ 3^m+1 &2& 4 & 10 & 28 &82 & ... \\ N & 1&2 & 1 & 2 & 1 & ... \end{array} $$ which can easily be extended to improve also intuition and finally might be proved by binomial expansion or by Fermat/Euler. So we find, that N is only 2 or 1 and that's the highest power of 2 occuring in $3^m+1$ - but clearly we can then have at most two solutions, because we have only 2 different N - that first two possibilites found in the table above prove to be true solutions.

Answer 2

$\mathsf{Hint}$: Consider mod 3, then mod 4, then mod 8.

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$\displaystyle\begin{array}{lllllll} \rm 2^n-3^m=1 & \implies & \rm (-1)^n\equiv 1\bmod 3 & \iff & \rm 2\mid n & \iff & \rm n=2u \\ \rm 4^u-3^m=1 & \implies & \rm (-1)^m\equiv-1\bmod 4 & \iff & \rm 2\nmid m & \iff & \rm m=2v+1 \\ \rm 4^u-3\cdot 9^v=1 & \implies & \rm 4^u\equiv 4\bmod 8 & \iff & \rm u=1 &\implies & \rm n=2,m=1\end{array}$

Answer 3

Hint:

1. When $n$ is odd, $2^n-1=(3-1)^n-1$ cannot be divided by $3$.
2. When $n$ is even, say $n=2k$, then $2^n-1=(2^k-1)(2^k+1)$. Note that $2^k-1$ and $2^k+1$ are coprime to each other.