Multiplying the first two equations and subtracting from the last, we get $y^{2023}z^2+z^{2023}y^2=0$. Hence either $y=0$ or $z=0$ or $y=-z$. But in the third case, $x=0$ and hence $y=z=0$. Hence the only solutions are $(x,x,0)$ and $(x,0,x)$ for arbitrary $x$.