The equation writes as $(p+q+1)(r-1) = pq$. Since $p+q+1 > \max\{p,q\}$, the only possibility is $r=2$ and $p+q+1 = pq$, i.e. $(p-1)(q-1) = 2$, with only solution(s) $\{p,q\} = \{2,3\}$.

Let wlog $p\ge q$. Then $r(p+q+1)=(p+1)(q+1)<p(p+q+1)\implies r<p$. Then taking $\mod p,p|(q+1)(r-1)\implies p|q+1\implies q+1\ge p\ge q$ which gives $p=3,q=2$.