If $\gcd(x,y)=1$, and $x^2 + y^2$ is a perfect sixth power, then $xy$ is a multiple of $11$ This is a problem that I don't know how to solve:

Let $x, y, z$ integer numbers such that $x$ and $y$ are relatively primes and  $x^2+y^2=z^6$ . Show that $x\cdot y$ is a multiple of $11$.


It's not complete answer
By computer search, it seems to be valid for co-prime $x$ and $y$.
Now, I'm trying to give the general solution of $(x,y,z)$.
\begin{align*}
  (a+bi)^3 &= a(a^2-3b^2)+b(3a^2-b^2)i \\[7pt]
  (a^2+b^2)^3 &=
  \underbrace{a^2(a^2-3b^2)^2}_{\Large{m^2}}+
  \underbrace{b^2(3a^2-b^2)^2}_{\Large{n^2}} \\[7pt]
  \begin{pmatrix} x \\ y \\ z \end{pmatrix} &=
  \begin{pmatrix}
    m^2-n^2 \\
    2mn \\
    \sqrt[3]{m^2+n^2}
  \end{pmatrix} \\[7pt] &=
  \begin{pmatrix}
    (a^2-b^2)(a^2+4ab+b^2)(a^2-4ab+b^2) \\
    2ab(a^2-3b^2)(3a^2-b^2) \\
    a^2+b^2 \end{pmatrix}
\end{align*}
Take absolute value if the solution is limited to positive.
The lowest non-trivial solution is $(117,44,5)$
