The answer is all $a, b, c, p$ with $\gcd(a, p) = 1$, $b^2 \equiv 3ac \pmod p$ and $p \equiv 2 \pmod 3$. It is easily checked that $a \not \equiv 0 \pmod p$. Note that the problem is equivalent to the statement: if $m, n \in \mathbb{F}_p$ satisfy $p \mid a(m^2+mn+n^2)+b(m+n)+c$ then $m = n$. Substituting $x = m+n$ and $y = m-n$ this becomes equivalent to having if $x, y \in \mathbb{F}_p$ satisfy $3ax^2+ay^2+4bx+4c \equiv 0 \pmod p$, then $y = 0$. Make the substitutions $u = 3ax+2b$ and $v = ay$ (all modulo $p$). The problem after some manipulations is now equivalent to: if $u, v \in \mathbb{F}_p$ satisfy $u^2+3v^2 \equiv C \pmod p$ where $C = 4(b^2-3ac)$ then $v = 0$. There are exactly $(p+1)/2$ quadratic residues modulo $p$ of the form $u^2$ and $(p-1)/2$ of the form $3v^2$ when $v \ne 0$. Cauchy Davenport then tells us that at least $\min \left(p, \frac{p+1}{2} + \frac{p+1}{2} -1 \right) = p-1$ residues are covered by $u^2+3v^2$ as $u, v$ vary over $\mathbb{F}_p$. As a result, at most one residue can possibly be missed, which must be exactly $C$. However scaling the equation by any perfect square also works, so $4C$, for instance should also be missed. As a result, $p \mid 4C-C$ which yields $C \equiv 0 \pmod p$. Finally to have $u^2+3v^2 \equiv 0 \pmod p$ force $u = v = 0$ we need $-3$ to be a quadratic non residue modulo $p$, or $p \equiv 2 \pmod 3$. Since all steps were reversible we have shown that the solution set is the same as the one described earlier.

If $p \equiv 1 \pmod 3$, Let $Q(x)=P(x)^{\frac{p-1}{3}}$. Then use our Lemma,$\sum_{x=0}^{p-1}{Q(x)} = \sum_{x=0}^{p-1}{P(x)^{\frac{p-1}{3}}} \equiv \sum_{x=0}^{p-1} {x}^{\frac{p-1}{3}} \equiv 0 \pmod p$. However, If we assume that $Q(x)=\sum_{i=\frac{p-1}{3}}^{p-1}{a_i x^i}$, $a_{p-1}=a^{\frac{p-1}{3}} \not \equiv 0 \pmod p$, use Lemma again: $\sum_{x=0}^{p-1}{Q(x)} = \sum_{x=0}^{p-1}{\sum_{i=\frac{p-1}{3}}^{p-1}{a_i x^i}} \equiv {\sum_{i=\frac{p-1}{3}}^{p-1}{a_i \sum_{x=0}^{p-1}{x^i}}} \equiv -a_{p-1} \not \equiv 0 \pmod p$. Contradiction.

Now let $Q(x)=P(x)^{\frac{p+1}{3}}=\sum_{i=\frac{p+1}{3}}^{p+1}{a_i x^i}$. With the same way in Step 1, we can get $0 \equiv \sum_{x=0}^{p-1}{Q(x)} = \sum_{x=0}^{p-1}{\sum_{i=\frac{p+1}{3}}^{p+1}{a_i x^i}} \equiv {\sum_{i=\frac{p+1}{3}}^{p+1}{a_i \sum_{x=0}^{p-1}{x^i}}} \equiv -a_{p-1} \pmod p\ \Rightarrow a_{p-1} \equiv 0 \pmod p$. As $(ax^3+bx^2+cx)^{\frac{p+1}{3}}=\sum_{i=\frac{p+1}{3}}^{p+1}{a_i x^i}$, we have $a_{p-1}=\frac{p+1}{3}a^{\frac{p-2}{3}}c+\frac{(p+1)(p-2)}{18}a^{\frac{p-5}{3}}b^2\equiv 0 \pmod p$, the same as $b^2 \equiv 3ac \pmod p$.