Solution:

We write $p+q^{2}=a^{2}$ with $a$ a positive integer. Then $p=a^{2}-q^{2}=(a-q)(a+q)$, and since $p$ is a prime number the factors $a-q$ and $a+q$ must be equal to $\pm 1$ or to $\pm p$. Since $a+q$ is a positive number, $a-q$ must also be positive, and moreover $a+q>a-q$, so the only possibility is $a+q=p,\ a-q=1$. It follows that $a=q+1$ and $p=a^{2}-q^{2}=2q+1$.

Suppose now, for contradiction, that $p^{2}+q^{n}$ is a square, $p^{2}+q^{n}=b^{2}$ with $b$ a positive integer. We have $q^{n}=b^{2}-p^{2}=(b-p)(b+p)$, and therefore, since the only divisors of $q^{n}$ are of the form $\pm q^{i}$ with $0 \leq i \leq n$, we must have $b-p=q^{i},\ b+p=q^{j}$ (as before it is easy to see that both factors are positive) and $q^{i} \cdot q^{j}=q^{n}$, that is, $i+j=n$; moreover, since $b-p<b+p$, we have $q^{i}<q^{j}$. Subtracting the equalities $b+p=q^{j},\ b-p=q^{i}$ term by term we obtain $2p=q^{j}-q^{i}=q^{i}\left(q^{j-i}-1\right)$.

We now distinguish the cases $i=0$ and $i>0$.

If $i=0$ we have $b-p=1$ and $b+p=q^{n}$: then $q^{n}=b+p=2p+1=4q+3$, so $3=q^{n}-4q=q\left(q^{n-1}-4\right)$. It follows that $q$ divides $3$, so $q=3$, and simplifying a factor of $3$ we arrive at the equation $3^{n-1}-4=1 \Rightarrow 3^{n-1}=5$, which has no integer solutions.

If instead $i>0$, from the equation $2p=q^{i}\left(q^{j-i}-1\right)$ we obtain $q|2p=2(2q+1) \Rightarrow q|2$, that is, $q=2,\ p=2q+1=5$, and $2p=q^{i}\left(q^{j-1}-1\right) \Rightarrow 10=2^{i}\left(2^{j-i}-1\right)$. It follows that $i$ is equal to $1$ (because $2^{i}$ divides $10$) and that $5=2^{j-1}-1$, but again this equation has no integer solutions.