In the course of answering another question here (link) I struggled quite hard at a rather unsatisfactory argument to show that $$x^6-y^6=z^2,$$ has no nontrivial integral solutions. For that question it was sufficient to assume $y$ even. From there is not much more work to prove it in full, though. It did leave me wondering whether there is a simpler argument. For completeness' sake I repeat my argument here:
Proposition: The Diophantine equation $x^6-y^6=z^2$ has no primitive integral solutions with $xyz\neq0$.
Proof. Let $x$, $y$ and $z$ be pairwise coprime integers such that $x^6-y^6=z^2$, with $|y|>0$ minimal. Then $(u,v,w)=(x^2,-y^2,z)$ satisfies $$u^3+v^3=w^2,$$ and it is a classical result by Euler that any primitive solution is of one of the following three forms: \begin{align} (1)&\begin{cases} u&=s(s+2t)(s^2-2st+4t^2)\\ v&=-4t(s-t)(s^2+st+t^2)\\ w&=\pm(s^2-2st-2t^2)(s^4+2s^3t+6s^2t^2-4st^3 + 4t^4) \end{cases} & 2\nmid s,\ 3\nmid(s-t),\\ \\ (2)&\begin{cases} u&=s^4-4s^3t-6s^2t^2-4st^3 + t^4\\ v&=2(s^4+2s^3t+2st^3+t^4)\\ w&=3(s-t)(s+t)(s^4+2s^3t+6s^2t^2+2st^3+t^4) \end{cases} &2\nmid(s-t),\ 3\nmid(s-t),\\ \\ (3)&\begin{cases} u&=-3s^4 + 6s^2t^2 + t^4\\ v&=3s^4+6s^2t^2-t^4\\ w&=6st(3s^4+t^4) \end{cases} &2\nmid(s-t),\ 3\nmid t, \end{align} after swapping $u$ and $v$ if necessary. In all three cases $s$ and $t$ are coprime integers. For a proof, see $\S 14.3.1$ of Henri's Cohen's Number Theory, Vol II, Analytic and Modern Tools.
If $y$ is even then $w=z$ is odd, so the solution $(u,v,w)=(x^2,-y^2,z)$ cannot be of the form $(3)$. It cannot be of the form $(2)$ because $-y^2\equiv0\pmod{4}$, whereas $u\equiv1\pmod{4}$ and $v\equiv2\pmod{4}$. It folows that it is of the form $(1)$, and so \begin{eqnarray} x^2&=&s(s+2t)(s^2-2st+4t^4)&=&s(s^3+8t^3),\\ y^2&=&4t(s-t)(s^2+st+t^2)&=&4t(s^3-t^3) \end{eqnarray} for two coprime integers $s$ and $t$ with $s$ odd and $3\nmid(s-t)$. The factors in each product are pairwise coprime, and hence they are all perfect squares, say $$s=S^2,\qquad t=T^2,\qquad s+2t=q^2,\qquad s^3-t^3=p^2,$$ for some integers $S$, $T$ , $p$ and $q$, not necessarily prime. Reducing mod $4$ shows that $t$ is even, and so we find $$S^6-T^6=p^2,$$ with $S$, $T$ and $p$ pairwise coprime and $T$ even, where clearly $|S|<|x|$ and $|T|<|y|$. This contradicts the minimality of $|y|$, and so we conclude that there does not exist any primitive solution with $y\neq0$ even.
If $y$ is odd then $z$ is even, and we have $$y^6+z^2=x^6,$$ which shows that $(y^3,z,x^3)$ is a primitive Pythagorean triple. This means there exist coprime integers $m$ and $n$ such that $$y^3=m^2-n^2,\qquad z=2mn,\qquad x^3=m^2+n^2,$$ because $z$ is even. Then $m+n=s^3$ and $m-n=t^3$ for two coprime integers $s$ and $t$, because $$y^3=m^2-n^2=(m+n)(m-n),$$ where $m+n$ and $m-n$ are coprime. It follows that $$2x^3=2(m^2+n^2)=(m+n)^2+(m-n)^2=(s^3)^2+(t^3)^2=(s^2)^3+(t^2)^3,$$ where $s$, $t$ and $x$ are pairwise coprime. But by the Main Theorem of this article no such solution exists, a contradiction.$\qquad\square$
