Croatia Mathematical Competitions

By moving $n^3$, we get a sum of cubes on the right-hand side:
$$p^m = n^3 + 8 = (n+2)(n^2 - 2n + 4).$$
Since $p$ is prime, each of the factors on the right-hand side must be a power of $p$:
$$n + 2 = p^{\alpha}, \quad n^2 - 2n + 4 = p^{\beta},$$
where $\alpha$ and $\beta$ are obviously positive integers.
Note that $n^2 - 2n + 4 \ge n + 2$, since that is equivalent to the claim that $n^2 - 3n + 2 \ge 0$, i.e. $(n-1)(n-2) \ge 0$, which holds because $n$ is a positive integer. Therefore, $\beta \ge \alpha$.
We can conclude that $p^\alpha$ divides both $n+2$ and $n^2 - 2n + 4$, so it also divides
$$n \cdot (n+2) - (n^2 - 2n + 4) = 4n - 4,$$
and then it also divides $4 \cdot (n+2) - (4n - 4) = 12$, so $p=2$ or $p=3$.
If $p=2$, then $n+2 = p^\alpha$ is at most 4, and since $n > 0$, it follows that $n=2$, which gives the solution $(2, 4, 2)$.
If $p=3$, then $n+2 = p^\alpha$ is 3, so $n=1$, which gives the second solution $(3, 2, 1)$.