We have given a prime $p$ and two non-negative distinct integers $x$ and $y$ satisfying the Diophantine equation $(1) \;\; x^4 - y^4 = (x^3 - y^3)p$. If $x=0$ or $y=0$, then $y=p$ or $x=p$ by equation (1) respectively. Next assume $x,y>0$. Equation (1)is equivalent to $(x - y)(x + y)(x^2 + y^2) = (x - y)(x^2 + xy + y^2)p$, yielding (since $x \neq y$) $(2) \;\; (x + y)(x^2 + y^2) = (x^2 + xy + y^2)p$. Set $d=GCD(x,y)$. Then there are two positive coprime distinct integers $s$ and $t$ s.t $(x,y) = (ds,dt)$, which inserted in equation (2) result in $(3) \;\; d(s + t)(s^2 + t^2) = (s^2 + st + t^2)p$. Combining equation (3) with the fact that $GCD((s + t)(s^2 + t^2),s^2 + st + t^2) = 1$, we obtain ${\textstyle \frac{d}{s^2 + st + t^2} = r \in \mathbb{N}}$, which inserted in equation (3) give us $(4) \;\; r(s + t)(s^2 + t^2) = p$. Seeing that $s^2 + t^2 > s + t > 1$ means $(s + t)(s^2 + t^2)$ is a composite positive integer, contradicting (since $p$ is a prime) equation (4). Consequently equation (1) is unsolvable when $xy \neq 0$. Conclusion: The only solutions of equation (1) are $(x,y) = (0,p)$ and $(x,y) = (p,0)$.

I have a bit different "solution'' from above, so I thought it would'nt be bad to share