South African Mathematical Olympiad, Third Round

Let the pair $(m, n)$ satisfy the equation. $m = 0$ implies that $2 = 19^n$, an impossibility. So we must have $m > 0$. Taking both sides modulo $10$ gives $1 \equiv (-1)^n \pmod{10}$, implying that $n$ must be even. Taking both sides modulo $20$ gives $-10m^2 + 1 \equiv (-1)^n \pmod{20}$, so that $2 \mid m^2$, whence $m$ is even too. So let us put $m = 2k$ and $n = 2l$ for a positive integer $k$ and a non-negative integer $l$. The original equation can now be written as
$$10m^2 - 1 = 20^{2k} - 19^{2l} = (20^k - 19^l)(20^k + 19^l).$$
As both $20^k + 19^l$ and $10m^2 - 1$ are positive (recall that $m \ge 1$), we must have that the integer $20^k - 19^l \ge 1$. It follows that
$$10m^2 - 1 = (20^k - 19^l)(20^k + 19^l) \ge 20^k + 19^l \ge 20^k + 1,$$
and we see that $20^k \le 10(2k)^2 - 2 = 40k^2 - 2$. This inequality is not true for $k = 2$, and it can be easily verified by induction that it also fails to hold for any $k > 2$. The inequality holds for $k = 1$, and it follows (and is easily checked) that the only pair that solves the original equation, is $(m, n) = (2, 2)$.