It follow from the Pytago solution. The equation has no solution.

It follows from the statement $ 0=0$, too... Seriously: the way you state works, but there still is a lot of ugly work to do. And the ugly (case)work is the main reason why I don't want to post a solution untill now... The idea is the same as for $ x^4+y^4=z^2$, but this one has a nasty second case.

Can someone please outline a proof? If it is that ugly indeed, I'll just remove it from PEN, or replace with its nicer brother.

let d=gcd(x,y), then $ d^4$ divides $ z^2$, then $ (x/d)^4 - (y/d)^4 = (z/d^2)^2$, if p prime divides two of the terms, divides the third, then p divides x/d and y/d(contradiction). So we can assume that x,y,z are coprimes each two $ (x^2 + z)(x^2 - z) = y^4$, then $ x^2 + z = m^4, x^2 - z = n^4$ $ 2x^2 = m^4 + n^4$ ;m>n. I will prove that this equation no has solution we suppose otherwise, then analagously above we can assume that 2x,m,n are coprimes each two, then 16 divides $ m^4 - 1 + n^4 - 1$, then x is odd we take the solution with x minimal m=n+2k, since m+n is even. then $ x^2 = (n^2 + 2nk + 4k^2)^2 - 8k^4$ $ 2k^4 = (A + x)/2 (A - x)/2$; since A and x are odd, if p prime divides (A+x)/2 and (A-x)/2, p divides A,x(p is odd), therefore p divides k p divides A then p divides n, then p divides m(contradiction) Case 1: $ (A + x)/2 = 2u^4, (A - x)/2 = v^4$ $ x = 2u^4 - v^4$. note that $ x^2 > (3k^2)^2 - 8k^4 = k^4$, then $ A^2 < 9x^2$,therefore A<3x, then u>v note that $ m^2 - mn + n^2 = A = 2u^4 + v^4, (m - n)/2 = k = uv$; u and v coprimes $ [(m + n)/2]^2 = m^2 - mn + n^2 - 3[(m - n)/2]^2 = 2u^4 - 3u^2v^2 + v^4 = (2u^2 - v^2)(u^2 - v^2)$; u>v $ gcd(2u^2 - v^2,u^2 - v^2) = gcd(u^2,u^2 - v^2) = gcd(u^2,v^2) = 1$ then $ 2u^2 - v^2 = r^2, w^2 = u^2 - v^2 = (u + v)(u - v)$ then $ u + v = s^2,u - v = t^2$ (s-t is even) or $ u + v = 2s^2,u - v = 2t^2$ $ r^2 = 2[(s^2 + t^2)/2]^2 - [(s^2 - t^2)]^2$ or $ r^2 = 2(s^2 + t^2)^2 - (s^2 - t^2)^2$ $ 2(r/2)^2 = [(s + t)/2]^4 + [(s - t)/2]^4 or 2r^2 = (s + t)^4 + (s - t)^4$ but $ x = 2u^4 - v^4 > 2u^4 - 2u^2v^2 + v^4 > (2u^2 - v^2)(u^2 - v^2) > 2u^2 - v^2 = r^2$ then x>r,r/2(contradiction) Case 2: $ (A + x)/2 = u^4, (A - x)/2 = 2v^4$ $ u^4 > 2v^4 > v^4$, then u>v $ m^2 - mn + n^2=A=u^4 + 2v^4, (m - n)/2 = k = uv$; u and v coprimes $ [(m + n)/2]^2 = m^2 - mn + n^2 - 3[(m - n)/2]^2 = u^4 - 3u^2v^2 + 2v^4 = (u^2 - 2v^2)(u^2 - v^2)$; $ gcd(2v^2 - u^2,v^2 - u^2) = gcd(v^2,v^2 - u^2) = gcd(v^2,u^2) = 1$ then $ u^2 - 2v^2 = r^2, w^2 = u^2 - v^2 = (u + v)(u - v)$ then $ 4u + v = s^2,u - v = t^2$ (s-t is even) or $ u + v = 2s^2,u - v = 2t^2$; s and t coprimes $ 4r^2 = - s^4 + 6s^2t^2 - t^4$ or $ r^2 = - s^4 + 6s^2t^2 - t^4$ in both cases $ R^2 = - s^4 + 6s^2t^2 - t^4 = 4s^2t^2 - (s^2 - t^2)^2$, note that R,s,t are coprimes each two $ (s^2 - t^2)^2 = (2st - R)(2st + R)$ ;gcd(2st-R,2st+R)=1,2,4 then $ 2st - R = a^2,2st + R = b^2$ or $ 2st - R = 2a^2,2st + R = 2b^2$ $ 4(s^2 + t^2)^2 = 4(8s^2t^2 - R^2) = 2(a^2 + b^2)^2 - (a^2 - b^2)^2$ or $ 4[2(a^2 + b^2)^2 - (a^2 - b^2)^2]$ then $ c^2 = 2(a^2 + b^2)^2 - (a^2 - b^2)^2$ then $ 2c^2 = (a + b)^4 + (a - b)^4$ $ c \le 2(s^2 + t^2) \le 4u$ $ 2x=2u^4 - 4v^4=2u^4 - (u^2 - r^2)^2 \ge 2u^4 - (u^2 - 1)^2 = u^4 + 2u^2 - 1 > u^4 \ge 8u \ge 2c$ ; if $ u \ge 2$ this is true since $ u^2=2v^2+r^2>1$. then x>c(contradiction).

Here you've got a nicer approach, which makes your statement no ugly at all: Lemma. There are no positive integers $ a, b, c, d$ for which $ a^2 - b^2 = c^2$ and $ a^2 + b^2 = d^2$ both hold. Proof. Assume the contrary and let us consider one solution $ (a, b, c, d)$ for which the minimum of $ b$ is attained. Summing yields $ 2a^2 = c^2 + d^2$, or equivalently, since $ 2 | c \pm d$, $ \left( \frac {d - c}{2} \right)^2 + \left( \frac {c + d}{2} \right)^2 = a^2$. It is easy to check that if $ \frac {d - c}{2}$ and $ \frac {c + d}{2}$ have any common factor, this leads to a solution with smaller components, fact which contradicts the minimality of $ b$. Hence they are coprime, and by the Pythagorean equation there exist relatively prime $ m, n \in \mathbb{N}$ such that $ d - c = 2(m^2 - n^2)$, $ c + d = 4mn$ and $ a = m^2 + n^2$. Hence $ b^2 = 4mn(m - n)(m + n)$ with all $ m, n, m - n, m + n$ coprime. Thus there are some positive integers $ x, y, z, t$ for which $ m = x^2$, $ n = y^2$, $ m - n = x^2 - y^2 = z^2$ and $ m + n = x^2 + y^2 = t^2$ and $ y^2 = n < 4mn(m^2 - n^2) = b^2$, so that $ (x, y, z, t)$ is a solution with $ y < b$, a contradiction. The case $ d - c = 4mn$ and $ c + d = 2(m^2 - n^2)$ follows the same reasoning. The proof of the proposed problem is now instantaneous: consider the solution with minimal $ z$ and notice that $ x, y$ need to be coprime. Hence either $ (x^2 - y^2, x^2 + y^2) = 1$, which in the view of $ (x^2 - y^2)(x^2 + y^2) = z^2$ shows that there exist some positive integers $ a, b$ such that $ x^2 - y^2 = a^2$ and $ x^2 + y^2 = b^2$, contradiction by the Lemma above, or $ (x^2 - y^2, x^2 + y^2) = 2$, which yields $ ((x^2 - y^2)/2)((x^2 + y^2)/2) = (z/2)^2$, and thus there are $ a, b \in \mathbb{N}$ for which $ x^2 - y^2 = 2a^2$ and $ x^2 + y^2 = 2b^2$, so $ a^2 + b^2 = x^2$ and $ b^2 - a^2 = y^2$, again a contradiction. Hence there are no solutions neither to this equation nor to the well-known $ x^4 + y^4 = z^4$.