Let $ n = \frac {x^2 + y}{xy + 1} > 1$, then $ x(x - ny) = n - y$ or $ y = \frac {x^2 - n}{nx - 1}$. From last we get $ x > n$. If $ y < n\to x > ny$ we get contadition. If $ y > n$ equation $ x(ny - x) = y - n,n < x < ny$ had not integer solution (as $ y-n=x(ny-x)\ge ny-1$ and thus $ 1-n\ge (n-1)y$). Therefore if $ n > 1$ we get $ y = n,x = n^2$ unique solution. But $ n = 1$ had many solutions $ (x,y) = (1,1),(t,t + 1)$.

Rust wrote: Let $ n = \frac {x^2 + y}{xy + 1} > 1$, then $ x(x - ny) = n - y$ or $ y = \frac {x^2 - n}{nx - 1}$. From last we get $ x > n$. Other method: From : $ y = \frac {x^2 - n}{nx - 1}$ The equation has solution if and only if $ nx - 1|x^2 - n$ $ \Longleftrightarrow nx - 1|n^2(x^2 - n)$ $ \Longleftrightarrow nx - 1|n^3 - 1$ Case 1:$ n > 1$ $ \Longrightarrow nx - 1 = \frac {n^3 - 1}{d}$ $ \Longrightarrow nx = \frac {n^3 - 1 + d}{d}$ Imply that $ n|d - 1$ If $ d\geq n + 1$ then $ x < n$ (tradition!) So $ d = 1$ Imply that $ x = n^2,y = n$ Case 2 $ n = 1$ $ (y,x) = (t + 1,t)$

we can consider the discriminant we have $ \frac {x^{2} + y}{xy + 1} = k \implies x^2 + y - kxy - k = 0$ $ \Delta = k^2y^2 - 4(y - k) = s^2$. the case $ k = y$ is obvious. we have two cases: 1-$ y > k$ which implies $ k^2y^2 > s^2 \implies (ky - 1)^2 \ge s^2 = k^2y^2 - 4y + 4k \implies - 2ky + 1 \ge - 4y + 4k \implies 4y - 4k + 1 \ge 2yk$ the rest is done with inequality. 2-$ y < k$ which implies $ k^2y^2 < s^2 \implies (ky + 1)^2 \le s^2 = k^2y^2 - 4y + 4k \implies 4yk + 1 \le - 4y + 4k$. the rest is also trivial we r done!