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.

Updated
We can try to find solutions ourself.

Equation 1.
$\quad x^4+x^2y^2+y^4 = z^2.$
Let 
$$\gcd(x,y) = m,\quad x=mX,\quad y=mY,\quad z=m^2Z,$$
then 
$$X^4+X^2Y^2+Y^2=Z^2,\quad\gcd(X,Y)=1.$$
$$(X^2+XY+Y^2)(X^2-XY+Y^2)=Z^2,\quad\gcd(X,Y)=1.$$
Multipliers in the left part are the same parity. If they are even, then both $X$ and $Y$ are even and coprime, but this conditions are incompatible. So the multipliers are odd, and if they can have only odd common divider, which are the divider of expressions 
$$3(X^2+XY+Y^2)-(X^2-XY+Y^2) = 2(X+Y)^2$$
and
$$3(X^2-XY+Y^2)-(X^2+XY+Y^2) = 2(X-Y)^2,$$
so numbers $X+Y$ and $X-Y$ share a common odd divider. And this odd divider must divide both their sum $2X$ and difference $2Y$, when $X$ and $Y$ are coprime. This contradiction proves that
$$\gcd(X^2-XY+Y^2,X^2+XY+Y^2)=1.$$
Product of coprime positive factors is the square of an integer. Consequently, each of them is the square of coprime integers:
$$\begin{cases}
X^2+XY+Y^2=U^2,\\
X^2-XY+Y^2=V^2,\\
\gcd(X,Y)=1,\\
\gcd(U,V)=1.
\end{cases}$$
So
$$\begin{cases}
X^2+Y^2=\dfrac{U^2+V^2}2,\\
2XY= U^2-V^2.
\end{cases}$$
Right part of the second equality is even, so $U$ and $V$ has the same parity and are odd. Then, $XY$ is even, when $X$ and $Y$ are coprime. 
The task is symmetric in the variables $X$ and $Y$, let $X$ is odd and $Y$ is even:
$$X\cdot\frac Y2 = \dfrac{U+V}2\frac{U-V}2.$$
Note than $\dfrac{U+V}2$ and $\dfrac{U-V}2$ are coprime, because any their common divider also divides their sum $U$ and difference $V$.
The last equation has а solution
$$X=ac,\quad\dfrac Y2=bd,\quad \dfrac{U+V}2=ad,\quad\dfrac{U-V}2=bc,$$
so
$$X=ac,\quad Y=2bd,\quad U=ad+bc,\quad V=ad-bc,\quad Z=\pm(a^2d^2-b^2c^2),$$
where 
$$\gcd(ac,2bd)=\gcd(ad,bc)=1.$$
Thus, $a$ and $c$ are odd and $a$, $b$, $c$, $d$ are pairwise coprime numbers.
That numbers must satisfy to equation
$$(ac)^2+4(bd)^2=(ad)^2+(bc)^2,$$
or
$$a^2(c^2-d^2)=b^2(c^2-4d^2).$$
$$\dots$$
Equation 2.
If 
$$3(xy)^2+z^2=(x^2+y^2)^2,$$
then
$$
\begin{cases}
xy=2m-1,\\
z=\dfrac{3x^2y^2-1}2,\\
x^2+y^2=\dfrac{3x^2y^2+1}2.
\end{cases}
$$
So
$$(xy+1)(3xy+1)=2(x+y)^2.$$
Then you can use the fact that equation
$$mn=p^2$$
has solutions
$$m=a^2c,\quad n=b^2c,\quad p=abc,$$
for $$m\in\left\{\dfrac{xy+1}2,xy+1\right\}$$
