The condition expands to $4q - 1 = 4r^2 - (4p^2 + 4p + 1) = (2r - 2p - 1)(2r + 2p + 1)$. So we must have $2r - 2p - 1 = 1$, which means $r = p + 1$. Thus $r = 3, p = 2$, which means $4q - 1 = 11$, so $(p, q, r) = (2, 3, 3)$, which works.

The only solution is $(2,3,3)$, which clearly works. Now, note that $p + q = r^2 - p^2$, so $q = r^2 - p^2 - p$, meaning \[4q - 1 = 4r^2 - (4p^2 + p + 1) = (2r)^2 - (2p + 1)^2 = (2r - 2p - 1) (2r + 2p + 1) \] Since $4q - 1$ is prime, one of $2r - 2p - 1, 2r + 2p + 1$ has absolute value $1$. Clearly $2r + 2p + 1 > 1$, so $|2r - 2p - 1| = 1$. If $2r - 2p - 1 = - 1$, we have $r = p$, but going back to the original equation, this implies $p + q = 0$, absurd. Hence $2r - 2p - 1 = 1$, so $r - p = 1$, implying $p = 2$ and $r = 3$. Now plugging back into the equation gives $q = 3$, as desired.