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How do you prove that the only integer solutions of $2^m+1=3^n$ are at $m=1, n=1$ and $m=3, n=2$? I have observed the following:

$9\equiv1$ (mod $4$), so any $n>1$ that is part of an integer solution must be even.

$3^n-1=2^m$, so $2^{m-1}=\sum_{k=0}^{n-1}3^k$.

$8\equiv-1$ (mod $9$), so any $m>1$ that is part of an integer solution must be odd divisible by 3.

However, none of these observations have gotten me anywhere.

I have no idea how to continue and would appreciate any help you can give. It would also be interesting to know if there are ways to solve more generalized versions of these kinds of problems (such as finding integer solutions to $2^m+1=k^n$, $2^m+k=3^n$, etc.).

InanimateBeing
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ASDF
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If $n = 2k$, you can write $2^m = (3^k-1)(3^k+1).$ These last two factors have to be powers of $2$ which differ by $2$. So you must have $3^k-1 =2$ and $3^k+1 = 4$. So $k=1$ and $n=2$.

B. Goddard
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