For any $x\ge 2$, note that $x^{n+1}-x^{n}-1\ge x^{n} > x$, and hence we may set $y=x^{n+1}-x^{n}-1$. Then $y+x^n = x^{n+1}-1$, and $x+y^n = ((x^{n+1}-1)-x^n)^n+x$. To show $x^{n+1}-1 | ((x^{n+1}-1)-x^n)^n+x$, we expand the right hand side and eliminate the terms divisible by $x^{n+1}-1$. We are left to show that $x^{n+1}-1 | x^{n^2}-x$, which is true, as $x^{n^2}-x = x(x^{n+1}-1)(1+x^{n+1}+...+x^{n^2-x-2})$