Test B Find all solutions in the set of natural numbers for the equation $2^{m}-3^{n}=1$.
(Former Soviet Union Mathematical Competition)
Solution: It is clear that the original equation has a solution $m=2, n=1$. We will prove that this is the only solution.
Assume $m>2$, we will discuss $n$:
(1) If $n$ is even, then $3^{n}=3^{2 k}=9^{k} \equiv 1(\bmod 8)$, so there is no solution when $n$ is even.
(2) If $n$ is odd and $n>1$,
$$
3^{2 k+1}+1=9^{k} \cdot 3+1 \equiv 3+1 \equiv 4(\bmod 8).
$$

While $2^{m} \equiv 0(\bmod 8)$, so there is no solution when $n$ is an odd number greater than 1.
Therefore, the equation has only the solution $m=2, n=1$.