= Solution
The <discriminant of a binary quadratic form> is $d=b^2-4ac$; it is unchanged by a <unimodular matrix> change of integral variables and satisfies $d\equiv0$ or $1\pmod4$. For such a $d$, the <discriminant criterion for prime representation by a binary quadratic form> says that some integral <binary quadratic form> of <discriminant of a binary quadratic form> $d$ represents an odd <prime> $p$ exactly when $d$ is a square modulo $p$, including zero. Equivalently,
$$
\boxed{\text{there is an integer }B\text{ with }B^2\equiv d\pmod{4p}.}
$$
These versions agree: a square root modulo $p$ can be chosen even when $d\equiv0\pmod4$ and odd when $d\equiv1\pmod4$, by adding $p$ if needed, and then the <Chinese remainder theorem> applies.
For completeness, if $f(x,y)=p$, then $(x,y)$ is primitive, since any common divisor would have its square divide $p$. Complete $(x,y)$ to a basis of $\mathbb Z^2$ with <determinant> one using <Bézout's identity>. The transformed <binary quadratic form> has leading coefficient $p$ and <discriminant of a binary quadratic form> $B^2-4pC=d$. Conversely, a suitable $B$ makes $[p,B,(B^2-d)/(4p)]$ an integral <binary quadratic form> representing $p$ at $(1,0)$.
For $x^2+3y^2$, the <discriminant of a binary quadratic form> is $-12$. The <prime> $3$ is represented. If $p\ne3$ is represented, reducing modulo $3$ gives $p\equiv x^2\equiv1\pmod3$, which also excludes $p=2$. To prove sufficiency for $p\equiv1\pmod3$, <quadratic reciprocity> gives
$$
\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac3p\right)=\left(\frac p3\right)=1.
$$
The criterion constructs a positive definite <binary quadratic form> $[p,B,C]$ of <discriminant of a binary quadratic form> $-12$. It is primitive: a common divisor of its coefficients would divide $p$, and if that divisor were $p$, then $p^2$ would divide $12$, impossible for $p\ne2,3$.
We must still establish that this form is equivalent to $x^2+3y^2$, rather than merely some form of the same <discriminant of a binary quadratic form>. In a positive definite integral <binary quadratic form>, replace $x$ by $x+ky$ to arrange $|b|\leq a$. If $c<a$, apply $(x,y)\mapsto(-y,x)$ and repeat. Each such exchange strictly decreases the positive integer leading coefficient, so this <reduction algorithm for a positive definite binary quadratic form> terminates with $|b|\leq a\leq c$. Then $12=4ac-b^2\geq3a^2$, so $a\leq2$. With $a=1$, parity forces $b=0$, giving $c=3$. With $a=2$, the only integral choices are $b=\pm2,c=2$, and these are not primitive. Thus every primitive positive definite <binary quadratic form> of <discriminant of a binary quadratic form> $-12$ is equivalent to $[1,0,3]$, and preserves its represented integers. Therefore
$$
\boxed{p=x^2+3y^2\text{ for integers }x,y\quad\Longleftrightarrow\quad p=3\text{ or }p\equiv1\pmod3.}
$$
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