Number Theory at its best!

$x^{2}+4y^{2}-2xy-2x-4y-8=(x-y-1)^{2}+3(y-1)^{2}-12$

$\Rightarrow(x-y-1)^{2}+3(y-1)^{2}-12=0$

$\Rightarrow(x-y-1)^{2}+3(y-1)^{2}=12$

As sum of squaring terms can't exceed 12 as R.H.S=12

$\Rightarrow-2\leq(y-1) \leq2$

$\Rightarrow-1\leq y\leq3$

As y is an integer

$\Rightarrow y\in {-1,0,1,2,3}$

Now check for each value of y in the above set such that x becomes an integer, we get;

$(x,y)\equiv(6,2),(0,2),(4,0),(-2,0),(4,3),(0,-1)$ Therefore, the answer should be $\boxed{6}$