LHS Is Large

$x^4+y^4=4xy-2$

Adding $-2x^2y^2$

$x^4+y^4-2x^2y^2=-2x^2y^2+4xy-2$

$(x^2-y^2)^2=-2(xy-1)^2$

or $(x^2-y^2)^2+2(xy-1)^2=0$

Sum of 2 squares is 0. So both squares are 0.

$x^2-y^2=0$ and $xy-1=0$

$x=\pm y$ and $xy=1$

So $(x=1,y=1)$ and $(x=-1,y=-1)$ are the 2 solutions

$\large x^4 + y^4 = 4xy - 2$ can be written as

$x^4 + y^4 + 2 = 4xy \\ \text{Also by AM- GM , we can say} \\ \dfrac{x^4+y^4 + 1 +1 }{4} \ge \large \sqrt[4]{x^4y^4} \\ \implies x^4+y^4 +2 \ge 4xy$

equality comes at $x = y = \pm 1$ .Hence only $\boxed{2}$ real solutions possible as $4xy$ is maximum.

Applying A.M.-G.M. inequality :

$x^4+y^4 \geq 2 \sqrt{x^4y^4}=2x^2y^2$

$2x^2y^2 \leq 4xy-2 \\ \implies x^2y^2-2xy+1 \leq 0 \\ \implies (xy-1)^2 \leq 0$

A perfect square cannt be less than zero.So, it is equal to zero.

$(xy-1)^2=0 \\ \implies xy-1=0 \\ \implies xy=1$ .

From this the suitible solutions are $(1,1);(-1,-1)$

By observing the equation, it's obvious that $4xy$ is the largest, so that limits down to $(1,1)$ and $(-1,-1)$ .