Determine all triplets $(p, n, k)$ of strictly positive integers such that $p$ is prime and satisfy the following equation:

$$
144 + p^n = k^2
$$
The number 144 being a square, we promptly factorize the equation. The equation rewrites as $p^{n}=k^{2}-144=k^{2}-12^{2}=(k-12)(k+12)$. In particular, since $(k-12)(k+12)=p^{n}$ and $k+12>0$, we have $k-12>0$, and $k+12$ and $k-12$ are powers of $p$. Let $a, b$ be positive integers such that $p^{a}=k-12$ and $p^{b}=k+12$. Since $k+12>k-12$, we have $b>a$, and by multiplying we get $a+b=n$.

If $a=0$, then $k=12+p^{a}=13$, and $p^{b}=13+12=25=5^{2}$, so $p=5$ and $b=2$. In particular, $n=a+b=2$, so $(p, n, k)=(5,2,13)$.

Otherwise, we have $a>0$. Note that $p^{a}\left(p^{b-a}-1\right)=p^{b}-p^{a}=k+12-(k-12)=24$. In particular, $p$ divides the left-hand side, so $p$ divides 24, hence $p=2$ or 3.

If $p=2$, we have $2^{a}\left(2^{b-a}-1\right)=2^{3} \times 3$. Since $2^{b-a}-1$ is odd, we must have $2^{a}=8$, so $a=3$. Thus $2^{b-a}-1=3$, so $2^{b-3}=4$, so $b-3=2$, so $b=5$. Thus $n=a+b=8$, and $k=12+p^{a}=12+2^{3}=20$. Therefore, $(p, n, k)=(2,8,20)$.

If $p=3$, we have $3^{a}\left(3^{b-a}-1\right)=2^{3} \times 3$. We cannot have $a \geqslant 2$, otherwise 9 would divide 24, so $a=1$. Thus $3^{b-a}-1=8$, so $3^{b-1}=9$, so $b-1=2$, so $b=3$. Thus $n=a+b=4$ and $k=12+p^{a}=15$. Therefore, $(p, n, k)=(3,4,15)$.

The three pairs found are solutions because $144+5^{2}=169=13^{2}$, $144+2^{8}=144+256=400=20^{2}$, and $144+3^{4}=144+81=225=15^{2}$. The solution pairs are therefore $(5,2,13)$, $(2,8,20)$, and $(3,4,15)$.