It suffices to consider non-negative $x, y, z$. Assume that there is a non-trivial solution, i.e. at least one of the numbers $x, y, z$ is positive. We pick such a solution with minimal sum $x + y + z$.
Now, $x^{2022}$ must be even since the other two terms are even, so we can write
$x = 2u$ and divide the new equation across by $2$ to get
$$2^{2021}u^{2022} + 10y^{2022} = 11z^{2022}.$$
As before, we conclude that $z = 2w$ is even and divide the new equation by $2$ to get
$$2^{2020}u^{2022} + 5y^{2022} = 11 \cdot 2^{2021}w^{2022}.$$
As before, we conclude that $y = 2v$ is even and divide the new equation by $2^{2020}$ to get
$$u^{2022} + 20v^{2022} = 22w^{2022}.$$
This is our original equation with $x, y, z$ replaced by $u, v, w$ whose sum is half the original sum. This gives a contradiction so there are no solutions.