Solution of problem T6/363 in M&Y

Find all $(x,y)\in\mathbb{Z}^2$ such that $x^{2007}=y^{2007}-y^{1338}-y^{669}+2$.

My solution.

Put $y^{669}=z$ then equation has form of $x^{2007}=z^3-z^2-z+2$(*). Assume that $(x,z)$ is an integer root of (*).

If $z=0$ then $x^{2007}=2$, contradiction!

If $z=-1$ then $x^{2007}=1$ , so $x=1$ and $y=-1$.

If $z\in\mathbb{Z}-\{-1,0\}$ then $(z+1)^3-(z^3-z^2-z+2)=4z^2+4z-1=(2z+1)^2-2>0$ and $(z^3-z^2-z+2)-(z-1)^3=2z^2-4z+3=2(z-1)^2+1>0$, therefore $(z-1)^3<(x^{669})^3<(z+1)^3$, or $x^{669}=z$. From (*) we have $-z^2-z+2=0$, or $z\in\{1,-2\}$ , therefore $x=1,y=1$.

Final, set roots of that equation is $\{(1,-1),(1,1)\}$.