Showing $(3 + \sqrt{3})$ is not a prime ideal in $\mathbb{Z}[\sqrt{3}]$

In addition to what was already said in the comments, I think there is another possibly faster way worth mentioning, which is using the following criterion for some Ideal $I$ being prime in a ring $R$ (which results immediately from the definitions):

$I\subset R$ is prime if and only if $R/I$ is an integral domain.

Using this, you would get

$$\mathbb{Z}[\sqrt{3}]/(3+\sqrt{3})\simeq(\mathbb{Z}[X]/(X^2-3))/(3+[X])\simeq \mathbb{Z}[X]/(X+3,X^2-3)\simeq\mathbb{Z}/(6),$$

which is certainly no integral domain, so $(3+\sqrt{3})\subset\mathbb{Z}[\sqrt{3}]$ is not prime.

(Whether or not this version is really shorter than the one you presented does however certainly depends on the amount of detail you would like to add to the isomorphisms used above...)

Why not write $3+\sqrt3=\sqrt3(1+\sqrt3)$?