Let $P(x,y,z)=x^3+y^3+z^3+mxyz=\prod\limits_{i=1}^3(\alpha_ix+\beta_iy+\gamma_iz+\delta_i)$. By homogenity, $\delta_1=\delta_2=\delta_3=0$. All other coefficients are nonzero, since if without loss of generality $\alpha_1=0$, $P(x,0,0)=0=x^3$ for all $x$. Hence, by rescaling the coefficients, we can assume $\alpha_1=\alpha_2=\alpha_3=1$. Now $x^3+y^3=P(x,y,0)=(x+\beta_1y)(x+\beta_2y)(x+\beta_3y)$. By comparing coefficients, $\beta_1+\beta_2+\beta_3=\beta_1\beta_2+\beta_2\beta_3+\beta_3\beta_1=0$ and $\beta_1\beta_2\beta_3=1$, so $\beta_1,\beta_2,\beta_3$ are roots of $t^3-1$, i.e. $\{\beta_1,\beta_2,\beta_3\}=\{1,\varepsilon,\varepsilon^2\}$ for $\varepsilon=\mathrm e^{\frac{2\pi\mathrm i}3}$. Analogously, $\{\gamma_1,\gamma_2,\gamma_3\}=\{1,\varepsilon,\varepsilon^2\}$. Now trying out the six permutations of the $\gamma_i$ shows that only $(x+y+z)(x+\varepsilon y+\varepsilon^2 z)(x+\varepsilon^2 y+\varepsilon z),(x+y+\varepsilon z)(x+\varepsilon y+z)(x+\varepsilon^2y+\varepsilon^2z),(x+y+\varepsilon^2z)(x+\varepsilon y+\varepsilon z)(x+\varepsilon^2y+z)$ are valid factorizations, leading to $m\in\{-3,-3\varepsilon,-3\varepsilon^2\}$.