\[ ab + 1 | a^2 + ab \]\[ ab + 1 | a + b \]Thus, $ab + 1 \le a + b$ WLOG $ a \le b$, Therefore, $ab < ab + 1 \le a + b \le 2a $, which means $a \le 1$. Therefore, since $a$ is a positive integer, $a = 1$. Now, since $a = 1$, then we can take any integer $b$. Check to the original equation, we will get $b + 1 | 0$, which is true for any positive integers $b$. Similarly, $(a,1)$ works too as $a + 1 | a^2 - 1$, works for any integers $a$. Therefore, $(1,k)$ and $(k,1)$ works for any positive integers $k$.