Local solutions of a Diophantine equation

I would like to tell you about my approach to this problem; clearly enough, it can easily be adapted to yield an even more general result.

Let $p \equiv 1 \, \, (\mathrm{mod} \, \, 3)$ be a prime number and let us denote with $\mathbf{J}$ the number of solutions of the congruence $3x^{3}+4y^{3}+5y^{3} \equiv 0 \, \, (\mathrm{mod} \, \, p)$. Then , by basic properties of the complex exponential function it follows that

\begin{eqnarray*} \mathbf{J} &=& \frac{1}{p} \sum_{x=0}^{p-1} \sum_{y=0}^{p-1} \sum_{z=0}^{p-1} \sum_{\lambda=0}^{p-1} e^{2\pi i \frac{3x^{3}+4y^{3}+5z^{3}}{p} \lambda} \\ &=& \frac{1}{p} \sum_{\lambda=0}^{p-1} \left(\sum_{x=0}^{p-1}e^{2\pi i \frac{3\lambda}{p}x^{3}}\right) \left(\sum_{y=0}^{p-1}e^{2\pi i \frac{4 \lambda}{p}y^{3}}\right) \left(\sum_{z=0}^{p-1}e^{2\pi i \frac{5 \lambda}{p}z^{3}}\right)\\ &=& p^{2} + \frac{1}{p} \sum_{\lambda=1}^{p-1} \left(\sum_{x=0}^{p-1}e^{2\pi i \frac{3\lambda}{p}x^{3}}\right) \left(\sum_{y=0}^{p-1}e^{2\pi i \frac{4 \lambda}{p}y^{3}}\right) \left(\sum_{z=0}^{p-1}e^{2\pi i \frac{5 \lambda}{p}z^{3}}\right). \end{eqnarray*} Appealing to the well-known estimate $ \left| \sum_{x=0}^{p-1} e^{2\pi i \frac{\omega}{p}x^{3}}\right| \leq 2 \sqrt{p}$ (which is valid for every $\omega$ coprime with $p$), we obtain that $$ \mathbf{J} > p^{2}-8p^{1.5}.$$ Since $p^{2}-8p^{1.5}>1$ for every $p \geq 65$, the problem has been reduced to verifying that the given congruence has a non-trivial solution for every $p \in E:=\{7, 13, 19, 31, 37, 43, 61\}$. I list below a non-trivial solution for the corresponding congruence for every $p \in E$ (even though at first sight they may strike you as having been chosen without rhyme or reason, they all were obtained by resorting to an idea of the Princeps Mathematicorum):

\begin{eqnarray*} p&=&07: \quad (x=3, y=6, z=0)\\ p&=&13: \quad (x=1, y=4, z=5)\\ p&=&19: \quad (x=13,y=0, z=3)\\ p&=&31: \quad (x=3, y=2, z=3)\\ p&=&37: \quad (x=1, y=4, z=0)\\ p&=&43: \quad (x=25,y=0, z=42)\\ p&=&61: \quad (x=46,y=32, z=41)\\ \end{eqnarray*} QED.