Problem 1303

Since $3^m - 2 \equiv 0 \pmod{7}$, one may easily show that $m \equiv 2 \pmod{6}$. If $m = 2$ then $n = 1$, which is a solution of the equation.

Assume that $m = 2s \ge 4$. Note that $2 + 7^n$ is divisible by $27$ in this case and
$$7^9 \equiv 1 \pmod{27}, \quad 7^1 \equiv -2 \pmod{27}.$$
Hence we have $n \equiv 4 \pmod{9}$. If we let $n = 9t + 4$, then
$$2 + 7^{9t+4} \equiv 2 + 7^4 \equiv 35 \pmod{37}.$$
For any positive integer $u$ less than $10$,
$$9^n \equiv 9, 7, 26, 12, 34, 10, 16, 33, 1 \pmod{37}.$$
Therefore there does not exist a solution for any $m \ge 4$. $\square$