Saudi Arabian Mathematical Competitions

Considering the equation $\bmod\ 5$ we get
$$3^{m} \equiv -1 \pmod{5},$$
so $m = 4k + 2$ for some positive integer $k$. Then, considering the equation $\bmod\ 7$ we get
$$-2^{n} - 9^{2k+1} \equiv 5 \pmod{7},$$
which means
$$2^{n} + 2^{2k+1} \equiv 2 \pmod{7}.$$
Since $2^{s} \equiv 1, 2, 4 \pmod{7}$ so the only possibility is $2^{n} \equiv 2^{2k+1} \equiv 1 \pmod{7}$ so $3 \mid n$ and $3 \mid 2k+1$. From the last one we get $3 \mid m$ so we can write $n = 3x$ and $m = 3y$. Therefore, the given equation takes the form $5^{3} \cdot 2^{3x} - 3^{3y} = 271$, or
$$\left(5 \cdot 2^{x} - 3^{y}\right)\left(25 \cdot 2^{2x} + 5 \cdot 2^{x} \cdot 3^{y} + 3^{2y}\right) = 271.$$
It follows that $25 \cdot 2^{2x} + 5 \cdot 2^{x} \cdot 3^{y} + 3^{2y} \leq 271$ and so $x < 2$. We conclude $x = 1$ and then $y = 2$. So $m = 6$ and $n = 3$.
$\square$