How to solve this for integers $(x^2 + 4)(y^2 + 1) = 8xy$

Question

How to solve the following equality $$(x^2 + 4)(y^2 + 1) = 8xy$$ for $x,y$ integers?

Answer 1

HINT:

$$(x^2+4)(y^2+1)-8xy=(xy-2)^2+(x-2y)^2$$

Answer 2

hint: $x^2+ 4 \ge 4x, y^2 + 1 \ge 2y$

Answer 3

HINT: solving for $y $ we get

$$y=\dfrac {4 x \pm \sqrt{-( x^2-4)^2}}{4 + x^2}$$

Now check which are the possible values of $x $, considering that we are looking for integer solutions.