How can find all solution of $x^2(y-1)+y^2(x-1)=1$ such that $x,y \in \mathbb{Z}$? 
How can find all solution of $x^2(y-1)+y^2(x-1)=1$ such that $x,y \in \mathbb{Z}$? 

I have no clue to find solution .
I tried to change into quadratic equation  $$y^2(x-1)+yx^2-(x^2+1)=0\\y=\frac{-y\pm\sqrt{x^4+4(x-1)(x^2+1)(-1)}}{2(x-1)}$$ 
but stop here ...  
Please help to me to find a method of solving . Thanks in advance.
Remark : by graphing that I found $$\quad{(x,y)=\\(2,1)\\(1,2)\\(2,-5)\\(-5,2)}$$

Write like this:
$$ xy (x+y)-(x+y)^2+2xy =1$$
If we put $a=x+y$ and $b=xy$ we get $$ b ={1+a^2\over a+2}\;\;\;\Longrightarrow \;\;\;a+2\mid a^2+1$$ 
Now since $a+2\mid a^2-4$ we have $a+2\mid 5$ so $a+2\in\{\pm 1,\pm 5\}$ and thus $a\in\{-3,-1,-7,3\}$
\begin{array}{c|c|c|c} 
a & b & x&y\\\hline
 -7  &  -10 &/&/\\
  -3 &  -10 &-5(2)&2(-5)\\
  -1 &  2 &/&/\\
   3&  2 &2(1)&1(2)\\
\end{array}
