The given equation can be written as
$$p^4 + p^3 + p^2 + p = q^2 + q \\ \Leftrightarrow p(p+1)(p^2+1) = q(q+1) \quad (1)$$
$$\Leftrightarrow p(p^2-1)(p^2+1) = q(q+1)(p-1) \quad (2)$$
For $q \le p$ is not possible. Hence we should have $q > p$.
Therefore from (2) we conclude that:
$$q|(p^2-1)(p^2+1). \quad (3)$$
Next we distinguish the cases::
* If $q \le p^2$, then $q^2 \le p^4$ and $q^2 + q \le p^4 + p^2 < p^4 + p^2 + p^3 + p$. Hence the equation has no solutions..
* If $q > p^2$, then $q \ge p^2 + 1$, and hence (3), since $q$ is prime, gives $q = p^2 + 1$. Thus from equation (1) we obtain
$$p(p+1) = q+1 \Leftrightarrow p^2 + p = p^2 + 2 \Leftrightarrow p = 2$$
Therefore, since $q = p^2 + 1 = 5$, we have the solution $(p,q) = (2,5)$.