I think of the elements of $\Bbb{Z}[x]/(p,x^2+3)$ as being expressions of the form $ax+b$, where $a$ and $b$ are in $\Bbb{Z}_p$ (or $\Bbb{Z}/p\Bbb{Z}$ if you prefer that cumbersome notation), and multiplication is subject to the condition $x^2 = -3$.
Since $\Bbb{Z}[x]/(p,x^2+3)$ is already a commutative ring, the only extra condition needed for it to be a field is that every nonzero element should have an inverse. So let $ax+b \in \Bbb{Z}[x]/(p,x^2+3)$, with $a$, $b$ not both zero. Note that if $a=0$ then this element certainly has an inverse, namely the element $b^{-1} \in \Bbb{Z}_p$. So assume that $a \ne0$.
The condition for $ax+b$ to have an inverse $cx+d$ is $1 = (ax+b)(cx+d) = (ad+bc)x + (-3ac + bd)$, giving the equations $ad+bc =0$ and $-3ac+bd = 1$. Solve these for $c$ and $d$ (in an elementary way as simultaneous equations) to get $c = -a(3a^2+b^2)^{-1}$, $d = b(3a^2+b^2)^{-1}$. The only thing that can go wrong there is if $3a^2+b^2$ is zero and so does not have an inverse.
Therefore the condition we are looking for is that $3a^2+b^2 \ne0$ or equivalently $\bigl( ba^{-1}\bigr)^2 \ne -3$. In other words, $-3$ should not be a quadratic residue mod $p$. In terms of Legendre symbols, this says that $$-1 = \Bigl(\frac{-3}p\Bigr) = \Bigl(\frac{-1}p\Bigr) \Bigl(\frac{3}p\Bigr).$$ I went to
https://en.wikipedia.org/wiki/Legendre_symbol to find that $$ \Bigl(\frac{-1}p\Bigr) = \begin{cases} 1&\text{if }\ p\equiv 1 \pmod 4, \\-1&\text{if }\ p\equiv 3 \pmod 4, \end{cases} \qquad \Bigl(\frac{3}p\Bigr) = \begin{cases} 1&\text{if }\ p\equiv 1 \text{ or }11 \pmod {12}, \\-1&\text{if }\ p\equiv 5 \text{ or }7 \pmod {12}. \end{cases}$$ Thus there are two possible cases:
Case 1. $p \equiv 1\pmod4$ and $p\equiv 5 \text{ or }7 \pmod {12}$. That gives $p\equiv 5 \pmod{12}$.
Case 2. $p \equiv 3\pmod4$ and $p\equiv 1 \text{ or }11 \pmod {12}$. That gives $p\equiv 11 \pmod{12}$.
Finally, those cases combine to give the answer $\boxed{p\equiv5 \pmod6}$.
The number of elements in that ring will always be $p^2.$