67th NMO Selection Tests for BMO and IMO

We show that $1$ is the only positive integer satisfying the condition in the statement. Clearly, $1=\frac{1^2+y}{1\cdot y+1}$ for any positive integer $y$, so $1$ is expressible in the required form for infinitely many pairs of positive integers.

Next, we prove that any integer $n\ge 2$ is uniquely expressible in the required form. Letting $x=n^2$ and $y=n$, it is readily checked that $\frac{x^2+y}{xy+1}=n$, so $n$ is indeed expressible in the required form.

To prove uniqueness, let $x$ and $y$ be positive integers such that $\frac{x^2+y}{xy+1}=n$. Alternatively, but equivalently, $x^2-nxy+y-n=0$, so the discriminant $\Delta =n^2{y}^2+4n-4y$ is a perfect square. It is readily checked that $(ny-2{)}^2<\Delta <(ny+2{)}^2$, so $\Delta =(ny\pm 1{)}^2$ or $\Delta =n^2{y}^2$. The former case is ruled out by noticing that $\Delta -n^2{y}^2=4(n-y)$ is even, and the latter yields $y=n$, so $x=n^2$.