XXIX Rioplatense Mathematical Olympiad

We claim that if $n=23k+20$ with $k$ a non-negative integer then the equation has no solution over the integers. Let's assume that there is one and get a contradiction. Indeed, by Fermat's little theorem, $$x^2+y^{11}\equiv n+z^{2022!}\equiv 20+(0\text{ or }1)\equiv 20\text{ or }21(\mathrm{mod}23)$$ and hence $$x^2+(-1,0\text{ or }1)\equiv 20\text{ or }21(\mathrm{mod}23)$$ because $-1,0$ and $1$ are the only congruence classes modulo $23$ such that its squares are congruent to $0$ or $1$ modulo $23$. It follows that $$x^2\equiv 19,20,21\text{ or }22(\mathrm{mod}23)$$ which is a contradiction because neither of them is a quadratic residue as can be seen by direct inspection.