Let $x+1=ad, y+1=bd$ where $d=(x+1,y+1)$. Then we know that $ab \mid (ad-2)a^2+(bd-2)b^2$ and hence $a \mid bd-2=y-1$ and $b \mid ad-2=x-1$. But $d \mid x+1,y+1$ by definition and hence $ad \mid y^2-1, bd \mid x^2-1$ as desired.

https://artofproblemsolving.com/community/c728438h1707823_system_of_divisibles $\begin{cases}(x+1)(y+1)|(x+1)^2 (x-1)+(y+1)^2 (y-1)\\ (x+1)^2 (y+1)^2 |(x+1)^2 (x-1)(y+1)^2 (y-1)\end{cases} \Rightarrow \begin{cases}(x+1)(y+1)|(x+1)^2 (x-1)\\ (x+1)(y+1)|(y+1)^2 (y-1)\end{cases}$ $\Rightarrow \begin{cases}(y+1)|(x+1)(x-1)\\ (x+1)|(y+1)(y-1)\end{cases}$ as required

Set$\displaystyle \ a=\frac{x^{2} -1}{y+1} , b=\frac{y^{2} -1}{x+1}$. Then $\displaystyle a+b,ab\in \mathbb{Z},a,b\in \mathbb{Q}$. Consier polynomial $\displaystyle f( t) =( t-a)( t-b) =t^{2} -( a+b) t+ab\in \mathbb{Z}[ t]$. From rational root theorem, $\displaystyle f( t)$ has only integer roots. It follows that $\displaystyle a,b\in \mathbb{Z}$.