Try to find all positive integers $n$, such that there exist at least two pairs of positive integers $(x, y)$, satisfying
$$
n=\frac{x^{2}+y}{x y+1} .
$$
If $xy$, then
$$
\begin{aligned}
x y+1 \mid x^{2}+y & \Rightarrow x y+1 \mid x^{2} y+y^{2} \\
& \Rightarrow x y+1 \mid\left(x^{2} y+y^{2}\right)-x(x y+1) \\
& \Rightarrow x y+1 \mid y^{2}-x .
\end{aligned}
$$

But from $x y+1>x, x y+1>y^{2}+1>y^{2}$, we know $\left|y^{2}-x\right| \leqslant \max \left\{y^{2}, x\right\}<x y+1$, hence $y^{2}=x$, at this point $\frac{x^{2} y+y}{x y+1}=y$.

In summary, if $\frac{x^{2}+y}{x y+1}=1$, then $(x, y)=(t, t+1)$ or $(1,1)$ (here, $t \in \mathbf{N}^{*}$), if $\frac{x^{2}+y}{x y+1}=l$ $(l \geqslant 2)$, then $(x, y)=\left(l^{2}, l\right)$.
Therefore, the positive integer $n$ that satisfies the requirement is only 1.