Theoretical way to prove no positive integer $n$ exists such that $n+3$ and $n^2+3n+3$ are both perfect cubes.

I have to prove that for any positive integer $n$ at least one of $n+3$ and $n^2+3n+3$ is not a perfect cube. Is there a methodical way to solve this problem?

I managed to solve it by contradiction, trying to force it into some sort of algebraic identity.

The solution I found is as follows:

If $n+3$ and $(n^2+3n+3)$ are both cubes then the product is a cube. The product is $n^3+6n^2+12n+9=(n+2)^3+1$. There are no two positive cubes with distance one between them, this is the contradiction.

For some reason this seems rather artificial, it makes me think that if one wished to make a very hard problem like this then he could look for much more complicated identities, and then actually finding the contradiction would be a lot harder. Is there a "safer" way to tackle these problems, without having to rely on "luck". (Of course it is not 100% luck because practice makes you a lot faster at finding the way to put together the expression)

$n+3=x^3$, $n=x^3-3$.

$n^2+3n+3=(x^3-3)^2+3(x^3-3)+3=x^6-3x^3+3=(x^2-(1/x))^3+x^{-3}>(x^2-1)^3$, but also $x^6-3x^3+3<(x^2)^3$, so it's not a cube.