There isn't a pair of primes such that the equation holds. The first case scenario is when $q=r$, then we get that $r^2-3r-2=0$, but this quadratic doesn't have a integer zero, so we get no solution. The second case scenario is when we have the system: $$r \mid q^2-q-1$$$$q \mid 2r+1$$ Now let $2r+1=qt$, then we have that $r=\frac{qt-1}{2}$, then we must have that: $$qt-1 \mid 2q^2-2q-2$$which implies that: $$qt-1 \mid 2q-2-t$$ Now here we also differentiate $2$ cases. The first case is when $2q-2-t \geq 0$, then we must have that: $$q(t-2) \leq -1-t \leq 0$$thus we must have that $t-2 \leq 0$, which in turn means that $t \leq 2$. By checking when $t=2$ and $t=1$, we get that we don't have solution. The second case is when $2+t-2q \geq 0$, then we must have that: $$q(t+2) \leq 3+t$$but this implies that: $$q \leq 1+ \frac{1}{t+2} \leq \frac{4}{3}$$which implies that we don't have a solution.