We show that $1$ is the only positive integer satisfying the condition in the statement. Clearly, $1 = \dfrac{1^2 + y}{1 \cdot y + 1}$ 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 $\dfrac{x^2 + y}{xy + 1} = n$, so $n$ is indeed expressible in the required form.

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