$xy + 1 | (x+y)(xy-1) \implies xy+1 | (x+y)(xy+1) - (x+y)(xy-1) = 2(x+y)$. So, we need $xy+1 | 2(x+y)$. This means $xy + 1 \le 2(x+y) \implies (x-2)(y-2) \le 3$. Now, a casebash shows that the only possibilities are $(x,y) = (3,5), (3,4), (3,3)$. Substituting in the original equation, we see that only $(3,5)$ works. So, the only solution is $(x,y) = (3,5)$ Oops, thanks @below - I forgot about when $x = 1,2$. If $x = 2$, then we need $2y+1 | 2y + 4$ and so $2y + 1 | 3$, which is impossible for $y \ge x$ So, if $x = 1$, then we need $y-1$ to be prime so $(x,y) = (1,p+1)$ work too So, the solutions are $(x,y) = (3,5), (1,p+1)$ for some prime $p$

What about $(x,y)=(1,p+1)$ ?