First, we prove that $n$ is square-free. If $d^2$ divides $n$ for a positive integer $d > 1$, then $(d^2)^2+d^2+1$ would be a prime number. But $d^4+d^2+1 = (d^2-d+1)(d^2+d+1)$, with both factors larger than $1$, which is a contradiction.

Thus, $n = p_1 \cdot p_2 \cdot \dots \cdot p_s$, where $s \in \mathbb{N}$ and $p_1 < p_2 < \dots < p_s$ are prime numbers. Let $p > 5$ be a prime number. Then $p \equiv 1 \pmod{6}$ or $p \equiv 5 \pmod{6}$. If $p \equiv 1 \pmod{6}$, then $p^2 + p + 1 \equiv 3 \pmod{6}$, and $p^2 + p + 1 > 3$ is composite.

If $p \equiv 5 \pmod{6}$, then $p^2-p+1 \equiv 3 \pmod{6}$, and $p^2-p+1 > 3$ is composite.

In conclusion, the only prime factors of $n$ can be $2$ and $3$, so $n \in \{2, 3, 6\}$. It is easy to check that all these three numbers fulfill the given condition.