Show that $n + 2$ and $n^2 + n + 1$ cannot both be perfect cubes

Question: If $n$ is a nonnegative integer, prove that $n + 2$ and $n^2 + n + 1$ cannot both be perfect cubes.

Possible solution: Suppose $n+2$ and $n^2 + n + 1$ are perfect cubes, their product $(n+2)(n^2 + n + 1)$ must also be a perfect cube.

However, note that $(n+2)(n^2 + n + 1)=n^3 + 3n^2 + 3n + 2 = (n + 1)^3 + 1^3$

By Fermat's Last Theorem, $a^n + b^n \neq c^n $ if $a,b,c,n$ are positive integers and $n>2$, therefore $a^3 + b^3 \neq c^3$ and $(n + 1)^3 + 1^3$ cannot be a perfect cube (can't be expressed in the form $c^3$ where $c$ is a positive integer)

I'm looking for alternative methods of solution, and some verification that the above proof is correct.

Except for $-1$, $0$ and $1$, the distance between consecutive perfect cubes is always greater than one. This is enough to conclude that $(n+1)^3+1$ is not a perfect cube when $n$ is nonnegative. (No need to invoke Fermat.)