For which primes p does x³ + y³ = p z³ admit non-trivial integer solutions?

The equation $x^3 + y^3 = p z^3$ has non-trivial integer solutions precisely when $p \equiv 1 \pmod{3}$ or $p = 3$. Here's why. Factor over $\mathbb{Q}(\omega)$ where $\omega = e^{2\pi i/3}$ is a primitive cube root of unity: $$x^3 + y^3 = (x + y)(x + \omega y)(x + \omega^2 y).$$ In $\mathbb{Q}(\omega)$, which has class number $1$, we can apply unique factorization. The equation becomes $$(x + y)(x + \omega y)(x + \omega^2 y) = p z^3.$$ If $p \equiv 2 \pmod{3}$, then $p$ remains prime in $\mathbb{Q}(\omega)$ (it is inert). By comparing ideal factorizations, one finds that at least one of the factors on the left must contain $p$ to an exponent not divisible by $3$, forcing a contradiction unless $x = y = z = 0$. If $p = 3$, then $3 = -\omega^2(1 - \omega)^2$ ramifies, and parametric solutions exist. For example, $$1^3 + 1^3 = 3 \cdot 1^3 \quad \text{and} \quad 2^3 + 2^3 = 3 \cdot 2^3.$$ If $p \equiv 1 \pmod{3}$, then $p$ splits as $p = \pi \bar{\pi}$ in $\mathbb{Q}(\omega)$. This allows solutions, e.g., for $p = 7$: $$2^3 + (-1)^3 = 7 \cdot 1^3.$$ For $p = 13$: $$3^3 + (-2)^3 = 13 \cdot 1^3.$$ In general, solutions correspond to finding $\alpha \in \mathbb{Z}[\omega]$ such that $\alpha \bar{\alpha} = p$ and then setting $x + y\omega = \alpha^3$ or similar multiplicative relations. This is a special case of the theory of cubic forms and is related to the Hasse principle for cubic surfaces.