Find prime factorization in the ring $\mathbb{Z}\left[\frac{-1 + \sqrt{3}}{2}\right]$

Question

Find prime factorization of the number $13$ in the ring $\mathbb{Z}\left[\frac{-1 + \sqrt{3}}{2}\right]$

My progress:

Let $w = \frac{-1 + \sqrt{3}}{2}$ and let $N(z) = z \bar z$ be the norm function.

$N(a + bw) = a^2 - ab + b^2$

$13 = (a + bw)\overline{(a + bw)} = a^2 - ab + b^2$. Let's divide the equation by $4$:

$13 = \frac{(2a-b)^2 + 3b^2}{4} \Rightarrow (2a-b)^2 + 3b^2 = 52 \Rightarrow$

$2a-b = 5 , b = 3 \Rightarrow a = 4, b = 3$

Thus $13 = (4 + 3w)\overline{(4 + 3w)}$

My questions:

1. Is it correct?
2. How can one show that $4 + 3w$ is prime in $\mathbb{Z}[w]$?