Find all integers $m, n \geq 0$ such that $3^{m}-7^{n}=2$.
We verify that there is only one solution for $m \leq 2: (m, n) = (2, 1)$. We assume therefore that $m \geq 3$ and $n \geq 2$. We apply the method of Dan Schwarz by rewriting the equation in the form

$$
3^{2}\left(3^{m-2}-1\right)=7\left(7^{n-1}-1\right)
$$

Thus $3^{2} \mid 7^{n-1}-1$. The order of 7 modulo $3^{2}$ is 3. Therefore, $19\left|7^{3}-1\right| 7^{n-1}-1$, so $19 \mid 3^{m-2}-1$.

The order of 3 modulo 19 is 18. Therefore, $37\left|3^{18}-1\right| 3^{m-2}-1$, so $37 \mid 7^{n-1}-1$.

The order of 7 modulo 37 is 9. Therefore, $3^{3}\left|7^{9}-1\right| 7^{n-1}-1$, so $3^{3} \mid 3^{2}\left(3^{m-2}-1\right)$. This is absurd, so there are no other solutions.