Find all positive integers $n$ and prime numbers $p$ such that $p^{n}+144=m^{2}$.
Our favorite remarkable identity allows us to rewrite the problem as

$$
p^{n}=(m-12)(m+12)
$$

Since $p$ is prime, the two factors $m-12$ and $m+12$ must necessarily be powers of $p$: $m-12=p^{\alpha}$ and $m+12=p^{\beta}$, with $\beta>\alpha \geqslant 0$ and $\alpha+\beta=n$. We then notice that $p$ divides both $m+12$ and $m-12$. Therefore, $p$ divides their difference, which is 24. Thus, $p=2$ or $p=3$.

If $p=2, m+12$ and $m-12$ are two powers of 2 that are 24 apart, which only occurs for $m+12=32$ and $m-12=8$. If $p=3, m+12$ and $m-12$ are two powers of 3 that are 24 apart, which only occurs for $m+12=27$ and $m-12=3$.

Finally, the only solutions to the problem are $p=2, m=20, n=8$ and $p=3, m=15$, $n=4$.