Find an ideal I of $ \mathbb Z[\sqrt 3]$ generated by an integer prime (i.e. a prime number in $\mathbb Z$) such that $\mathbb Z[ \sqrt 3]/I$ is not an integral domain.
Thoughts: We need to find some prime number that generates an ideal. This ideal should have the property that upon constructing a quotient of its ring, it should exhibit zero divisors. If we choose the prime number to be $3$, we consider $I =(\{ 3\})$. Then observe that $$\sqrt{3}=0+ 1 \cdot \sqrt{3} \in \mathbb Z[\sqrt{3}]$$ And also that $\sqrt{3}$ is not divisible by $3$ in $\mathbb Z[ \sqrt 3]$, therefore its residue class modulo 3 will be $\overline{\sqrt{3}}$ $$ \overline{\sqrt{3}} \cdot \overline{\sqrt{3}} =3 \equiv 0 \bmod 3$$ So in $\mathbb Z[ \sqrt 3]/(3)$ we observe that there is at least one zero divisor and therefore it cannot be an integral domain.
I'm fairly new to integral domains and ideals, is this reasoning correct/on point?
Couldn't fit this in a comment so I will write it here. Your reasoning is correct.
Furhtermore for all primes $p\in\mathbb{Z}\setminus\{2,3\}$, let $\mathfrak{{p}}=(p)_{\mathbb{Z}[\sqrt{3}]}$.
Then $\mathfrak{p}$ is a prime ideal if and only if $x^2-3$ has no roots modulo $p$, in which case $\mathbb{Z}[\sqrt{3}]/\mathfrak{p}\cong \mathbb{F}_{p^2}$.
On the other hand $\mathfrak{p}$ is not prime if and only if $x^2-3$ has a root $r$ modulo $p$ (so it has two roots, $-r$ is the other one), and in this case $\mathbb{Z}[\sqrt{3}]/\mathfrak{p}\cong \mathbb{F}_p \times \mathbb{F}_p$.
When the prime $p=3$ you will have $\mathbb{Z}[\sqrt{3}]/\mathfrak{p}\cong \mathbb{F}_3[x]/(x^2)$.
When the prime $p=2$ you will have $\mathbb{Z}[\sqrt{3}]/\mathfrak{p}\cong \mathbb{F}_2[x]/((x-1)^2)\cong \mathbb{F}_2[x]/(x^2)$.
In other words $\mathbb{Z}[\sqrt{3}] = \mathbb{Z}[x]/(x^2-3)$ and $\mathbb{Z}[\sqrt{3}]/(p)_{\mathbb{Z}[\sqrt{3}]} \cong \mathbb{F}_p[x]/(x^2-3)$