Let $xy|(x^2+y^2+1)$. Prove that $\frac{x^2+y^2+1}{xy}=3$

Question

Let $x,y -$ positive integers, such that $xy|(x^2+y^2+1)$. Prove that $$\frac{x^2+y^2+1}{xy}=3.$$

My work so far:

1) If $x=y=1$ then $\frac{1^2+1^2+1}{1}=3$

2) Let $x=1$ (or $y=1$) $\frac{1^2+y^2+1}{y}=k \Rightarrow y^2-ky+2=0$

$D=k^2-8=m^2 \Rightarrow (k-m)(k+m)=8 \Rightarrow k=\pm3, m=\pm1 $, but $k\in \mathbb N \Rightarrow k=3$

3) $x,y>1$

$xy|(x^2+y^2+1) \Rightarrow x^2+y^2+1=kxy, k \in \mathbb N$

$D=k^2y^2-4y^2-4$

Here I need help...