If $\gcd(x,y)=1$, and $x^2 + y^2$ is a perfect sixth power, then $xy$ is a multiple of $11$

First, an observation: not both of $ x, y $ can be odd. Indeed, then $ x^2 + y^2 $ would be $ 2 $ modulo $ 4 $, so it wouldn't even be a perfect square, much less a perfect sixth power. In addition, clearly both of them cannot be even either, as then they would not be coprime. In the following, assume wlog that $ x $ is even and $ y $ is odd.

Pass to the ring $ \mathbf Z[i] $ of Gaussian integers. It is a principal ideal domain where $ 11 $ remains inert, and the quotient $ \mathbf Z[i]/(11) \cong \mathbb F_{11}(\sqrt{-1}) \cong \mathbb F_{121} $. Note that

$$ \gcd(x+yi, x-yi) = \gcd(x+yi, 2x) = \gcd(2yi, x - yi) $$

so that $ \gcd(x+yi, x-yi) $ divides $ \gcd(2x, 2y) = 2 $. However, the prime lying over $ 2 $ does not divide $ x+yi $, as it divides $ x $ and does not divide $ y $. It follows that $ x+yi $ and $ x-yi $ are coprime, and the factorization

$$ (x + yi)(x - yi) = z^6 $$

implies that $ x + yi = r^6 $ for a Gaussian integer $ r $. Clearly, then, $ x + yi $ is also a sixth power modulo $ 11 $ in $ \mathbf Z[i] $. $ x + yi $ is a unit modulo $ 11 $ since not both of $ x $ and $ y $ are divisible by $ 11 $, and since it is a sixth power in $ \mathbb F_{121}^{\times} $, which is a group of order $ 120 $, it has order dividing $ 20 $, so that it is a root of $ X^{20} - 1 $. This polynomial has exactly $ 20 $ roots in $ \mathbb F_{11}(\sqrt{-1}) $: any element $ a + b\sqrt{-1} $ with $ a, b \in \mathbb F_{11} $ and where exactly one of $ a, b $ is zero is a root, and there are no other roots, since a polynomial of degree $ n $ has at most $ n $ roots in a field. It follows that $ x + yi $ is congruent to $ a + b\sqrt{-1} $ where exactly one of $ a, b $ is zero modulo $ 11 $, which implies that one of $ x, y $ is zero modulo $ 11 $. Finally, note that if $ x/11 $ is a Gaussian integer and rational, it is an integer, so that either $ x $ or $ y $ is indeed divisible by $ 11 $ in $ \mathbf Z $.

NOTE This answer was written before the OP edited the question to constrain the integers as relatively prime.

I think the assertion is false. Consider the pythagorean triple $(3,4,5)$. Now multiply the triple by $5^2=25$, so we get the triple $(75,100,125)$, i.e.,

$$75^2+100^2=125^2 = 5^6$$

But $75 \cdot 100 = 7500$, which is not divisible by $11$.

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 $(117,44,5)$