Find all integral solutions for the Diophantine Equations $x^4 - x^2y^2 + y^4 = z^2$ and $x^4 + x^2y^2 + y^4 = z^2$. 
Find all integral solutions for the Diophantine Equations $$x^4 - x^2y^2 + y^4 = z^2$$ and $$x^4 + x^2y^2 + y^4 = z^2$$

I basically think that to solve these equations we need to use the fact that all Pythagoras Triplets are like $(k \cdot 2mn, k \cdot (m^2 - n^2), k \cdot (m^2 + n^2))$. The above equations can be modified a little bit to make all the terms perfect squares. Then the first equation would be $$(x^2 - y^2)^2  + (xy)^2 = z^2$$ and the second one would be $$z^2 + (xy)^2 = (x^2 + y^2)^2$$
I have found no other clues. Please help me proceed. 
Thanks.

COMMENT.- I feel but I have no proof that there are only the trivial solutions $(x,y,z)= (t,0,t^2),(0,t,t^2)$. I give here an outline of what could perhaps lead to a proof.
$$x^4-x^2y^2+y^4=z^2\iff (x^2-y^2)^2+x^2y^2=z^2\qquad (1)$$ Hence, as it is well known, 
$$\begin{cases}x^2-y^2=t^2-s^2\\xy=2ts\\z=t^2+s^2\end{cases}\qquad (2)$$
It follows in particular $$x^2+s^2=t^2+y^2\qquad (3)$$ The general solution of $(3)$, which likely as for Pithagorean triples comes from an identity easily verified, is given by
$$\begin{cases}2x=aX+bY\\2s=aY-bX\\2t=aX-bY\\2y=aY+bX\end{cases}\qquad (4)$$
where $a,b$ are arbitrary integers and $X,Y$ two parameters. 
In any case, because of $(3)$,  $x$ and $y$ must have the form, given by $(4)$, for certain integer values $a,b,X,Y$. However, in order to use parameters $t$ and $s$ we need besides to fit our values with the constraint $xy=2ts$ so we get the condition
$$(aX+bY)(aY+bX)=2(aX-bY)(aY-bX)\iff 3ab(X^2+Y^2)=(a^2+b^2)XY$$
It follows $$\left(\frac XY\right)^2-\left(\frac{a^2+b^2}{3ab}\right)\left(\frac XY\right)+1=0$$
Hence $$ \frac{6abX}{Y}= a^2+b^2\pm \sqrt{ (a^2+b^2)^2-(6ab)^2}$$ so the equation $$(a^2+b^2)^2-(6ab)^2=c^2\iff (a^2+b^2+6ab)(a^2+b^2-6ab)=c^2$$
It could be useful perhaps for someone. I stop here the comment.
