Solution:

The possible values of $(m, n, p)$ are $(13,2,5)$, $(20,8,2)$ and $(15,4,3)$.

We rewrite the equation in the form $p^{n}=m^{2}-144=(m+12)(m-12)$. Since the only divisors of $p^{n}$ are powers of $p$, the equation is equivalent to the system
$$\left\{ \begin{array}{l} m+12=p^{a} \\ m-12=p^{b} \end{array} \right.$$
where $a$ and $b$ are natural numbers such that $a>b \geq 0$ and $a+b=n$. Subtracting the second equation from the first, we obtain
$$24=p^{b}\left(p^{a-b}-1\right) .$$
If $b=0$, this last equation becomes $24=p^{a}-1$, from which we immediately obtain the solution $p=5, n=2, m=13$.

If instead $b>0$, since $p^{b}$ divides $24=2^{3} \cdot 3$, then $p=2$ or $p=3$. Moreover, since $p$ does not divide $p^{a-b}-1$, $p^{b}=8$ or $p^{b}=3$. Substituting, we find the other two solutions $p=2, n=8, m=20$ and $p=3, n=4, m=15$.