Solution:
Let $p(x) \equiv a x^3 + b x^2 + c x + d$. Finding the solutions with $x = y$ is obvious, just solve the cubic $a x^3 + b x^2 + (c - 1)x + d = 0$.

For $x \neq y$, we have $x - y = a(y^3 - x^3) + b(y^2 - x^2) + c(y - x)$.

Dividing by $y - x$ gives $a(x^2 + x y + y^2) + b(x + y) + c + 1 = 0$.

Put $s = x + y$, $t = x y$ and this becomes $a s^2 - a t + b s + c + 1 = 0$ (*).

We also have $x + y = a(x + y)(x^2 - x y + y^2) + b(x^2 + y^2) + c(x + y) + 2d$, or $s = a s (s^2 - 3 t) + b(s^2 - 2 t) + c s + 3 d$.

Substituting for $t$ from (*) we get a cubic in $s$. Solving, we then recover $t$ from (*) and then solve a quadratic to get $x, y$ from $s, t$.