Mod $23$ will work easily

math.geniuz.2004 wrote: Let $n > N$ be a positive integer and let $(x,y,z)$ be a solution to the equation $x^2 + y^{11} - z^{2022!} = n$. Then $$x^2 + y^{11} - z^{2022!} = n = (x+y^{11}-z^{2022!})^2.$$ Why? math.geniuz.2004 wrote: $$x^2 + y^{11} - z^{2022!} = n = (x+y^{11}-z^{2022!})^2.$$ Expanding the right-hand side, we get $$x^2 + y^{11} - z^{2022!} = x^2 + 2xy^{11} - 2xz^{2022!} + y^{22} - 2y^{11}z^{2022!} + z^{4044}.$$ Equating the coefficients on both sides, we get \begin{align*} 1 &= 2xy^{11} - 2xz^{2022!}, \ 0 &= 2y^{11} - 2y^{11}z^{2022!}, \ -1 &= - 2y^{11}z^{2022!} + z^{4044}. \end{align*} Solving the equations above, we get $$y^{11} = z^{2022!} \quad \text{and} \quad x = z^{2022!}.$$ There are a lot of factorials missing in your computations. math.geniuz.2004 wrote: Therefore, $x^2 + y^{11} - z^{2022!} = (z^{2022!})^2 + (z^{2022!})^{11} - (z^{2022!})^{2022!} = z^{4044} + z^{22244} - z^{2022!}$. Since the right-hand side is an integer, the left-hand side must be an integer as well, which means that $n$ must be an integer. However, this contradicts the fact that $n > N$ The problem states that $n$ is an integer. I don't see why this contradicts the fact that $n>N$.

By Fermat, $t^{22}\equiv 1\pmod{23}$ if $23\nmid t$. So, $y^{11}\in\{-1,0,1\}\pmod{23}$ and $z^{2022!}\in\{0,1\}\pmod{23}$, yielding $y^{11}-z^{2022!}\in\{-2,-1,0,1\}\pmod{23}$. Now, the quadratic residues in mod 23 are $\{0,1,2,3,4,6,8,9,12,13,16,18\}$. Doing some messy math, we get that $n\equiv 20\pmod{23}$ is the only uncovered residue class, closing the case.

grupyorum wrote: By Fermat, $t^{22}\equiv 1\pmod{23}$ if $23\nmid t$. So, $y^{11}\in\{-1,0,1\}\pmod{23}$ and $z^{2022!}\in\{0,1\}\pmod{23}$, yielding $y^{11}-z^{2022!}\in\{-2,-1,0,1\}\pmod{23}$. Now, the quadratic residues in mod 23 are $\{0,1,2,3,4,6,8,9,12,13,16,18\}$. Doing some messy math, we get that $n\equiv 20\pmod{23}$ is the only uncovered residue class, closing the case. exactly.