A complete list of solutions is

Solution 1. Any factor of $n$ is a factor of $3p^2 + 1$. Hence, $3$ cannot be a factor of $n$. Hence, $p$, the smallest non-factor of $n$, can only be $2$ or $3$, as non-factors of $4$ and above cannot be the smallest non-factor. If $p = 2$, $n$ must be an odd factor of $3p^2 + 1 = 13$, i.e. $n = 1$ or $n = 13$. If $p = 3$, then $n$ must be an even factor of $3p^2 + 1 = 28$, i.e. $n = 2, 4, 14, 28$. This gives the solutions as tabulated above.

Solution 2. Clearly $p > 1$ as $1$ is a factor of all positive integers $n$. The minimality of $p$ implies that $p-1$ is a factor of $n$ which is a factor of $3p^2+1$, so $p-1 \mid 3p^2+1$. We also have $p-1 \mid 3p^2-3 = 3(p-1)(p+1)$. By subtraction $p-1 \mid 4$, so $p-1 \in \{1, 2, 4\}$ and $p \in \{2, 3, 5\}$ so $3p^2 + 1 \in \{13, 28, 76\}$. Checking the factors of $\{13, 28, 76\}$ for possible values of $n$ gives the solutions above.

Solution 3. Trying $p = 2, p = 3, p = 4$ and $p = 5$ and trying in turn each factor of $3p^2 + 1$ as a candidate for $n$ gives all the solutions stated. If $p > 5$ then $2, 3, 4, 5$ are factors of $n$ and in particular $n$ is multiple of $10$, hence $3p^2 + 1$ and $3(3p^2 + 1) = 9p^2 + 3$ as well, i.e. $-p^2 + 3 \equiv 9p^2 + 3 \equiv 0 \pmod{10}$. But this means that $3$ would have to be a quadratic residue modulo $10$, which is not the case.