I'm carrying out a congruence procedure, so that you have different approaches.
If $p \, | \, n^3 + 1$ and $p \, | \, n^2 + 2$, then $2n \equiv 1 \pmod p$, which means $ -8 \equiv 8n^3 = (2n)^3 \equiv 1^3 \equiv 1 \pmod p, $ hence $9 \equiv 0 \pmod p$ and assuming $p$ is a prime means $p = 3$. This means the $\gcd$ is a power of $3$. Now I'm checking powers of $3$ by hand using congruences.
EDIT : As miracle's comment says, I got far too carried away by liking primes so much : a good way to say that this proof is done is that $9 \equiv 0 \mod p$ means $p \, | \, 9$, hence getting examples is enough to get our answer.
If $n \equiv 0 \pmod 3$, this $\gcd$ is clearly $1$.
If $n \equiv 1 \pmod 3$, $n^3 + 1 \equiv 1 \pmod 3$ so that the $\gcd$ is again $1$.
If $n \equiv 2 \pmod 3$, then $9 \, | \, n^3 + 1$ and $3 \, | \, n^2 + 2$. But letting $n = 3k+2$ we notice that \begin{align*} (3k+2)^3 + 1 & = (3k)^3 + 3 \cdot (3k)^2 \cdot 2 + 3 \cdot 3k \cdot 2^2 + 2^3 +1 \\& \equiv 9(k+1)\pmod{27} \end{align*} which is $0$ if and only if $k \equiv 2 \pmod 3$. Carry on! We get $ (3k+2)^2 + 2 \equiv (3k)^2 + 2 \cdot (3k) \cdot 2 + 2^2 + 2 = 9k^2 + 12k + 6 \equiv 0 \pmod{27} $ if and only if $ 3k^2 + 4k + 2 \equiv 0 \pmod 9 $ but reading this $\bmod 3$, we get $k \equiv 1 \pmod 3$, a contradiction.
I must say it is a little longer than the $\gcd$ proof : I expected people to put up to $\gcd$ proof, so I've shown this one instead. I like those proofs because they're mechanical ; I didn't have to think much to write it, I just chose to go this way and things always go as expected (assuming the question asks something that's true, obviously)... Now I've only proven that the $\gcd$ divides $9$ : again, to show that the $3$ possibilities actually happen, pull off examples like Andre Nicolas.
Hope that helps,