Proving $x^4+y^4=z^2$ has no integer solutions

There are no positive integer solutions to $x^4+y^4=z^2.$ By contradiction suppose $(x,y,z)$ is a positive solution with the least possible $z.$

If a prime $p$ divides $x$ and $y$ then $p^4$ divides $z^2,$ so $p^2$ divides $z.$ But then $(x',y',z')=(x/p,y/p,z/p^2)$ is a positive solution with $z'<z,$ contradicting the minimality of $z.$

Therefore $x,y$ are co-prime. So $(x^2,y^2,z)$ is a Fundamental Pythagorean triplet. So there exist co-prime positive integers $m,n$, not both odd, with $\{x^2,y^2\}=\{m^2-n^2, 2mn\}$ and $z=m^2+n^2.$ WLOG $x^2=m^2-n^2$ and $y^2=2mn.$ Note that $x$ is odd and $y$ is even.

Now $x^2+n^2=m^2.$ A prime divisor of both $x$ and $n$ must divide $m,$ but $\gcd(m,n)=1.$ So $x,n$ are co-prime and $(x,n,m)$ is a Fundamental Pythagorean triplet. So there are co-prime positive integers $a,b$ with $\{a^2-b^2, 2ab\}=\{x,n\}$ and $a^2+b^2=m.$ Since $x$ is odd we have $x=a^2-b^2$ and $n=2ab.$

Now $y^2=2mn=2(a^2+b^2)(2ab)=4(a^2+b^2)ab.$ But $\gcd (a,b)=1$ so the members of $\{a,b, a^2+b^2\}$ are pair-wise co-prime and their product $(y/2)^2$ is a square. Therefore each of $a,b,a^2+b^2$ is a square: There are positive integers $d,e,f$ with $a=d^2,b=e^2,a^2+b^2=f^2.$ Therefore $$ f^2=a^2+b^2=d^4+e^4.$$ But $f^2=a^2+b^2=m<m^2+n^2=z,$ so $f\leq f^2<z.$ This contradicts the minimality of $z.$

The following is not an answer, almost like a lengthy version of a comment.

Say $a=x^2$ and $b=y^2$.

Then the equation becomes $a^2+b^2=z^2$.

Now we can conclude that $a,b,z$ will of the following forms courtesy this proof: $$a=d\cdot 2pq$$ $$b=d\cdot (p^2-q^2)$$ $$z=d\cdot (p^2+q^2)$$

where $d=\mathrm{gcd}(a,b,z)$ and $p,q \in \mathbb{N}$.

Now $a$ and $b$ are $2$ perfect squares.

So the problem now reduces to this:

Prove that numbers having the form $d\cdot 2pq$ and $d\cdot (p^2-q^2)$ cannot be perfect squares.

See if this helps.

P.S. If you still want some alternate method to find solutions of $a^2+b^2=z^2$, you can look up this link.

I think you equation has infinitely many integer solutions.

For example $\{(a,0,a^2)|a\in\mathbb Z\}$.