Solve the equation $3^{m}+7=2^{n}$ in integers.

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The left side is not less than 7, so $n$ is not less than 3. Then the right side is not less than 8, and $m$ is non-negative.

For $m=0$, we have $2^{n}=8$ and $n=3$.

If $m>0$, then modulo 3 the left side is congruent to 1, so $n$ is even, that is, $n=2k$. $2^{2k}-7=4^{k}-7 \equiv 1$ (mod 4), hence $m$ is also even: $m=2p$. Transform the equation:

$7=2^{2k}-3^{2p}=\left(2^{k}-3^{p}\right)\left(2^{k}+3^{p}\right)$.

From this, $2^{k}+3^{p}=7, 2^{k}-3^{p}=1$, and we get the unique solution $k=2, p=1$, that is, $m=2, n=4$.

## Answer

$(0,3),(2,4)$.