Assume $3 \not \mid x$. Show that one of the prime factors of $x^2 + 3$ is of the form $12n + 7$.
If prime $p$ is a factor of $x^2+3$ then $x^2 \equiv -3 \pmod p$. This means $(\frac{-3}{p}) = 1$.
If a prime factor is of the form $12n+7$, for some integer $n$, then it must be that $(\frac{-3}{12n+7}) = 1$. This is what we need to show.
$$ \begin{align} (\frac{-3}{12n+7}) & = (\frac{-1}{12n+7}) \times (\frac{3}{12n+7}) \\ \\ &= (-1) \times (\frac{3}{12n+7}) \tag{Proposition (7.11)} \\ \\ & = (-1) \times -(\frac{12n+7}{3}) \tag{Corollary (7.17)} \\ \\ & = (-1) \times -(\frac{1}{3}) \\ \\ &= -1 \times -1 = 1 \end{align} $$
And so a prime of the form $12n+7$ is a factor of $x^2+3$.
Note: Primes of a different form may also satisfy $(\frac{-3}{p}) = 1$, and therefore also be factors of $x^2+3$.