How to solve this diophantine equation: $x^2+2y^2=x^2y^2-2000$

Question

Solve the following diophantine equation: $$x^2+2y^2=x^2y^2-2000$$

I tried this by adding and subtracting terms, but so far, no avail.

$(\pm 2y^2+4xy, \pm2xy)$

I don't know how to start either. Please give a hint.

Answer 1

Tips:$$x^2+2y^2=x^2y^2-2000 \\ x^2y^2-x^2-2y^2=2000 \\ x^2y^2-x^2-2y^2+2=2002 \\ (x^2-2)(y^2-1)=2002$$

Answer 2

$$x^2=\dfrac{2(y^2-1+1001)}{y^2-1}=2+\dfrac{2002}{y^2-1}$$

$$y^2-1\le2002\iff y^2\le2003<45^2$$

$$\implies 2\le y\le44$$

Also $y^2-1$ must be odd and must divide $2002=2\cdot7\cdot11\cdot13$