Find all positive integers $m, n$ such that $3^{m}-2^{n}=1$.
If $n=0,1$ we find the solution $(1,1)$. Now $n \geqslant 2$ and modulo 4 we deduce that $m$ is even. Thus

$$
\left(3^{m^{\prime}}-1\right)\left(3^{m^{\prime}}+1\right)=3^{2 m^{\prime}}-1=2^{n}
$$

So $3^{m^{\prime}}-1 \mid 3^{m^{\prime}}+1$ and thus $3^{m^{\prime}}-1 \mid 2$. We deduce that $m^{\prime}=1$ and we find the solution $(2,3)$. Another solution using Zsigmondy: we rewrite the equation as $2^{n}+1=3^{m}$, and we eliminate the case $n=1$ which gives the solution $(1,1)$. The left-hand side always has a primitive factor different from $3=2+1$, except for the case $n=3$ which will give the solution $(2,3)$.