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

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

$$(x^2+4)(y^2+1)-8xy=(xy-2)^2+(x-2y)^2$$
hint: $x^2+ 4 \ge 4x, y^2 + 1 \ge 2y$
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.