T时 $\star \star$ Find all positive integer solutions $m, n$ for the equations (1) $2^{m}-3^{n}=1$, (2) $3^{n}-2^{m}=1$.

 untranslated part: T时 $\star \star$ 

The translated part: Find all positive integer solutions $m, n$ for the equations (1) $2^{m}-3^{n}=1$, (2) $3^{n}-2^{m}=1$.
Equation (1) has only one positive integer solution $m=2, n=1$. In fact, because $3^{2 k}+1=9^{k}+1=2(\bmod 8).3^{2 k-1}+1=4(\bmod 8)$, where $k$ is a positive integer. Therefore, $3^{n}+1$ (where $n$ is a positive integer) cannot be divisible by 8. Thus, if positive integers $m, n$ satisfy $2^{m}-3^{n}=1$, then $m \leqslant 2.2-3^{n}=1$ is impossible, so $2^{2}-3^{n}=1$. Hence, we get $m=2, n=1$.

Equation (2) has only two sets of positive integer solutions, namely $m=n=1$ and $m=3, n=2$. In fact, if $n=2 k+1, k$ is a positive integer, then $3^{n}=3^{2 k+1} \equiv 3(\bmod 4)$. So $2^{m}-3^{n}-1 \equiv 2(\bmod 4)$, thus $m=1$, yielding $n=1$. If $n=2 k, k$ is a positive integer, then $2^{m}=3^{2 k}-1=\left(3^{k}-1\right)\left(3^{k}+1\right)$, and $3^{k}-1,3^{k}+1$ are two consecutive even numbers, and their product is a power of 2, so they are 2 and 1, respectively. Thus, $k=1, n=2$, yielding $m=3$.