Determine all pairs $(m, n)$ of positive integers such that $2^{m}-1=3^{n}$
Let's look at the powers of 3. These are alternately $1, 3, 1, 3, 1, \ldots$ modulo 8, so they can only be 1 or 3. If $m \geqslant 3$, we have $8 \mid 2^{m}$ and $3^{n} \equiv 7 \pmod{8}$, which is impossible.

We must therefore have $m<3$.

If $m=2$ we have $3^{n}=3$ so $n=1$.

If $m=1$ we have $3^{n}=1$ so $n=0$.

If $m=0$ we have $3^{n}=0$ which is impossible.

Conversely, the pairs $(2,1)$ and $(1,0)$ are indeed solutions to this equation since $2^{2}-1=3^{1}$ and $2^{1}-1=3^{0}$.