Proving $x^4+y^4=z^2$ has no integer solutions I need solution check to see if I overlooked something:
If $x^4+y^4=z^2$ has an integer solution then $(\frac{x}{y})^4+1=(\frac{z}{y^2})^2$ has a solution in rationals. 
Second equation is equivalent to $X^4+1=Y^2$ which can be written as cubic Weierstrass form $$v^2=u^3-4u$$ where $u=\frac{2(Y+1)}{X^2}$ and $v=\frac{4(Y+1)}{X^3}$. 
By using SAGE we can see that Mordell-Weil group of this elliptic curve is empty (SAGE returns empty set, but I'm not sure if that is enough to conclude this), so we can conclude that there are no rational solutions to the second equation and no integer solution to the first one. 
edit: No integer solution in positive integers. 

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 $z.$
