Irish Mathematical Olympiad

Because $p^n = n^3 + 27 = (n+3)(n^2 - 3n + 9)$, there are positive integers $x, y$ such that $p^x = n+3$ and $p^y = n^2 - 3n + 9$. As $n^2 - 3n + 9 - (n+3) = n^2 - 4n + 6 = (n-2)^2 + 2 > 0$ we have $p^x < p^y$ and so $p^x|p^y$. Hence $(n+3)|(n^2 - 3n + 9)$. But $n^2 - 3n + 9 = (n+3)(n-6) + 27$, thus $(n+3)|27$, which gives $n+3 \in \{3, 9, 27\}$.
The case $n = 24$ is impossible, because $n^3+27 = 24^3+27 = 3^3(8^3+1) = 3^6 \cdot 19$ is not a power of a prime. For $n = 0$ and $n = 6$ we obtain the following solutions $(p, m, n)$: $(3, 3, 0)$ and $(3, 5, 6)$.