Note that $\Delta = p^2 - 4(q + 1) = s^2$ for some $s$ so $p^2 + q^2 = s^2 + (q + 2)^2$. Now we have two distinct representations as the sum of two squares so by Fermat it is composite. Unless $\pm p = q + 2$, but then it is in fact even, so composite since $2q^2 + 4q + 4 > 2$, with equality not holding because then $q = -1$ so zero is a root of the quadratic, a contradiction. Or $\pm q = q + 2$ which implies $q = -1$, which fails for the same reason. Done.