Can both $n+3\; \text{and}\; n^2+3$ both be cubic numbers at same time? 
Can both $n+3\; \text{and}\; n^2+3$ both be cubic number at same time? Where $n$ is an integer number. Not necessarily positive.    

I tried writing $x^3 = n+3$ and expressing $n^2+3$ in terms of $x$. I found $x^6 -6x^3+12$ but this doesn't help. How do I prove this?

Slightly overkill, but if $n + 3$ and $n^2 + 3$ are both cubes, then so is their product, and so
$$ (n + 3)(n^2 + 3) = n^3 + 3n^2 + 3n + 9 = (n + 1)^3 + 2^3 $$
would be a cube. But as is well known, the only possible solutions to this can occur when one of the cubes is $0$, and so we have that either $n = -3$ or $n = -1$, and we can verify that neither of these yield solutions.
