If $x,y \in \mathbb{Z}.$ Then the ordered pair of $(x,y)$ for which
$3x^4-2(19y+8)x^2+361y^2+2(100+y^4)+64=2(190y+2y^2)$
Try: From $$3x^4-2(19y+8)x^2++2y^4+357y^2-380y+264=0$$
For real roots, its discriminant always $\geq 0$
$$4(19y+8)^2-4\cdot 3 \cdot (2y^4+357y^2-380y+264)\geq 0$$
$$361y^2+64+304y-6y^4-1125y^2+1140y-792\geq 0$$
So $$6y^4+764y^2-1444y+728\leq 0$$
I am struck at that point. did not how to solve further
Completing a few squares shows that your equation can be written as $$((19y-(10+x^2))^2+2(y^2-1)^2+2(x^2-9)^2=0.$$ As all terms are nonnegative we find that $x=3$ and $y=1$.
At first I tried completing squares in a more roundabout way, yielding the expression $$2x^4+(x^2-19y-8)^2+2(y^2-1)^2+198=684y.$$ This immediately shows that $y>0$ and that $2(y^2-1)^2+198<684y$, which implies $y<7$. This leaves you with six quartics in $x$ with integer coefficients; you can check whether they have integer roots using the rational root theorem.