To find all prime numbers $p$ and nonnegative integers $x \neq y$ satisfying the equation:

$$x^4 - y^4 = p(x^3 - y^3),$$

we start by factoring the equation on both sides. Notice that:

$$x^4 - y^4 = (x^2 - y^2)(x^2 + y^2) = (x-y)(x+y)(x^2 + y^2),$$

and

$$x^3 - y^3 = (x-y)(x^2 + xy + y^2).$$

Substituting these factorizations into the given equation, we obtain:

$$(x-y)(x+y)(x^2 + y^2) = p(x-y)(x^2 + xy + y^2).$$

Assuming $x \neq y$, divide both sides by $x-y$ (since $x-y \neq 0$):

$$(x+y)(x^2 + y^2) = p(x^2 + xy + y^2).$$

Re-arrange the equation to:

$$x^3 + xy^2 + x^2y + y^3 = px^2 + pxy + py^2.$$

This simplifies to:

$$x^3 + xy^2 + x^2y + y^3 = px^2 + pxy + py^2.$$

Next, rearrange terms:

$$x^3 + xy^2 + x^2y + y^3 - px^2 - pxy - py^2 = 0,$$

or

$$x^3 + y^3 + xy^2 + x^2y - px^2 - py^2 - pxy = 0.$$

Assume $x = p$ and $y = 0$, checking these values in the original equation gives:

$$p^4 - 0^4 = p(p^3 - 0^3),$$

This simplifies to:

$$p^4 = p \cdot p^3,$$

which is always true.

Similarly, setting $y = p$ and $x = 0$ gives:

$$0^4 - p^4 = p(0^3 - p^3),$$

or

$$-p^4 = -p \cdot p^3,$$

which simplifies to:

$$p^4 = p \cdot p^3,$$

which is also true.

Thus, the only solutions are $(p, x, y) = (p, p, 0)$ and $(p, x, y) = (p, 0, p)$ for any prime $p$.

The solution set is:

$$\boxed{\{ (p, p, 0), (p, 0, p) \}}$$