A common divisor $d$ of the given numbers divides also $n^2 + 2n + 3 - (n^2 + 2) = 2n + 1$, $n \cdot (2n + 1) = 2n^2 + n$, $2n^2 + n - 2 \cdot (n^2 + 2) = n - 4$, $2n + 1 - 2 \cdot (n - 4) = 9$.
Therefore, if $d$ is a prime, then $d = 3$, and the only other prime factor contained by the given numbers is $2$, appearing in at most one of them. Moreover, at least one of the numbers has the exponent of $3$ at most $2$.
If $n$ is even, then $n^2 + 2$ is even, but not divisible by $4$ and the possible cases are:
I) $n^2 + 2 = 2 \cdot 3$;
II) $n^2 + 2 = 2 \cdot 3^2$;
III) $n^2 + 2n + 3 = 3$;
IV) $n^2 + 2n + 3 = 3^2$.
Only case II gives a solution, namely $n = 4$.
If $n$ is odd, then $n^2 + 2n + 3$ is even, but not divisible by $4$ and the possible cases are:
V) $n^2 + 2n + 3 = 2 \cdot 3$;
VI) $n^2 + 2n + 3 = 2 \cdot 3^2$;
VII) $n^2 + 2 = 3$;
VIII) $n^2 + 2 = 3^2$.
We get the solution $n = 1$ (corresponding to the cases V and VII).