IRL_ABooklet

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.