Find all $m, n \in \mathbb{N}$ such that $3^{m}-2^{n} \in\{-1,5,7\}$.
i) Case $3^{m}-2^{n}=-1$: we rewrite this as $3^{m}=2^{n}-1$, which will allow us to use Zsigmondy. The case $n=1$ gives us the solution $(0,1)$. We know that $2^{2}-1$ is divisible by 3 (we find the solution (1,2) here as well), so for $n \geqslant 3$ and except possibly for $n=6$, $2^{n}-1$ will have a prime factor different from 3, which is absurd. We verify that the case $n=6$ does not yield any solutions.

ii) Case $3^{m}-2^{n}=5$: we rewrite this as $3^{m}-1=2^{n}+4$ and use Zsigmondy. We check $n=0,1,2$ and find the solution $(2,2)$. For $n \geqslant 3$, looking modulo 4 we find that $m$ is even. We then use LTE in the case $p=2$ and $m$ even, and $v_{2}\left(3^{m}-1\right)=$ $v_{2}(3-1)+v_{2}(3+1)+v_{2}(m)-1 \geqslant 3$, which is absurd!

iii) Case $3^{m}-2^{n}=7$: we rewrite this as $3^{m}-1=2^{n}+8$. The cases $m=0,1,2$ give us $(2,1)$ as a solution. For $n \geqslant 3$, modulo 8 we get a contradiction.

In conclusion, the solutions are $(0,1),(1,2),(2,2)$ and $(2,1)$.