Notice that if $p>3$ is a prime, then $p^2\equiv 1 \pmod{24}$. Then we have that if $p>q>3$, then $p^2-q^2\equiv 0 \pmod{24}$, so $2|p^2-q^2$. Thus, since neither $p$ or $q$ can be even and $p>q$, then $q=3$. Then we have that $\frac{p^2-9}{4}$ is odd. Equivalently, we have that $p^2-9\equiv4 \pmod{8}\Rightarrow p^2\equiv 5 \pmod{8}$, contradiction, because $n^2\equiv 0, 1, 4 \pmod{8}$. Thus, there are no solutions.