Determine all solutions $(m, n) \in$ \textbf{N}^2 of the equation:

$$
3^{n}-2^{m}=1
$$
If $n$ is odd, $3^{n} \equiv 3(\bmod 8)$, so we must have $2^{m} \equiv 2(\bmod 8)$, which means $m=1$ and $n=1$. If $n=2 k, 2^{m}=3^{2 k}-1=\left(3^{k}+1\right)\left(3^{k}-1\right)$, this implies that $3^{k}+1$ and $3^{k}-1$ are two powers of 2, whose difference is $2$: this can only be 2 and 4, so $k=1, n=2, m=3$.