Solve in integers $y^3-1=x^4+x^2$

Solve in integers $y^3-1=x^4+x^2$ First you can add $1$ and factor to get $y^3=(x^2+x+1)(x^2-x+1)$ Since the GCD of $(x^2+x+1)$ and $(x^2-x+1)$ is $1$, then they both must be perfect cubes. I don't know what to do after that. Any help? Thanks

As you have noted, by rearranging the equation you can find that $y^3=(x^2+x+1)(x^2-x+1)$, and by considering the GCD of the two polynomials on the RHS you know that both must be perfect cubes.
Note that one solution is $(x,y)=(0,1)$.
By symmetry, we may assume that $x> 0$ for all other solutions.
Consider $(x^2+x+1)-(x^2-x+1)=2x<2(x^2+x+1)^{1/2}$. This says that the difference between our two cubes must be less than twice the square root of the larger cube. However, the difference between any two cubes $(n+k)^3-n^3>3n^2$ is larger than thrice the square of the cube root of the smaller cube.
Thus, $3(x^2-x+1)^{2/3}<2(x^2+x+1)^{1/2}$, which is a contradiction if $x>1$.
Therefore, the only possible solutions beside $(x,y)=(0,1)$ must have $x=\pm 1$, which obviously do not work.