I initially looked at this and believed it could be a fairly simple proof.
I started by stating that if $2$ is not a prime, then it can divide the product of $2$ elements of this ring, but cannot divide the individual elements.
It's easy enough to show it divides the product, as: $(1+\sqrt{-3})(1-\sqrt{-3}) = 4$ and it's clear that $2 \mid 4$.
Now to show it doesn't divide the individual elements I used the function:
$N: R \rightarrow \mathbb{Z}$: $(a + b\sqrt{-3}) \rightarrow (a^2 + 3b^2)$
and if $2$ does divide these elements then, taking $(1+\sqrt{-3})$
$(1+\sqrt{-3}) = 2a$ and therefore $N(2)N(a) = N(1+\sqrt{-3})$
so $4N(a) = 4$, therefore $a=1$ however even for the other element I found that $a=1$ and this shows that $2$ does divide the individual elements which can't happen if $2$ is prime? So I'm very unsure what I'm doing incorrectly.