This equation is unsolvable. Modulo 9 analysis shows that $z \ne 0 \pmod{3}$ and $z \ne \pm 1 \pmod{9}$. Let $p$ be a prime divisor of $z^2 + 3$ for which 3 is not a cubic residue. From $$z^2 + 3 = 1/3(3y^3 - (-3x)^3)$$ it follows that $3 \mid \nu_p(z^2 + 3)$. But we will show that $z^2 + 3$ ($z \ne 0 \pmod{3}$ and $z \ne \pm 1 \pmod{9}$) always has prime divisor $q$ for which 3 is a cubic nonresidue and $\nu_q(z^2 + 3)$ is not divisible by 3.

The proof requires the ring $\mathbb{Z}[\omega]$, $\omega = \frac{-1 + i\sqrt{3}}{2}$. $$z^2 + 3 = (z + 1 + 2\omega)(z - 1 - 2\omega)$$ We can calculate cubic character using the cubic reciprocity law. Let $z \equiv 1 \pmod{3}$, replace $z$ with $-z$ otherwise. \begin{equation*} \begin{split} \left(\frac{3}{z + 1 + 2\omega}\right)_3 = \left(\frac{3}{\omega(z + 1 + 2\omega)}\right)_3 = \\ = \left(\frac{3}{-2 + (z - 1)\omega}\right)_3 = \omega^{1/3(z - 1)} \ne 1 \\ \end{split} \end{equation*} Hence, $z + 1 + 2\omega$ has prime divisor $\pi$ such that 3 is a cubic nonresidue modulo $\pi$ and $\nu_p(z + 1 + 2\omega)$ is not divisible by 3. The same applies to complex conjugate $\overline{\pi} \mid z - 1 - 2\omega$. Thus, we can take $q = \pi \overline{\pi} \in \mathbb{Z} $.

This equation is unsolvable. Modulo 9 analysis shows that $z \ne 0 \pmod{3}$ and $z \ne \pm 1 \pmod{9}$. Let $p$ be a prime divisor of $z^2 + 3$ for which 3 is not a cubic residue. From $$z^2 + 3 = 1/3(3y^3 - (-3x)^3)$$ it follows that $3 \mid \nu_p(z^2 + 3)$. But we will show that $z^2 + 3$ ($z \ne 0 \pmod{3}$ and $z \ne \pm 1 \pmod{9}$) always has prime divisor $q$ for which 3 is a cubic nonresidue and $\nu_q(z^2 + 3)$ is not divisible by 3.

The proof requires the ring $\mathbb{Z}[\omega]$, $\omega = \frac{-1 + i\sqrt{3}}{2}$. $$z^2 + 3 = (z + 1 + 2\omega)(z - 1 - 2\omega)$$ We can calculate cubic character using the cubic reciprocity law. Let $z \equiv 1 \pmod{3}$, replace $z$ with $-z$ otherwise. \begin{equation*} \begin{split} \left(\frac{3}{z + 1 + 2\omega}\right)_3 = \left(\frac{3}{\omega(z + 1 + 2\omega)}\right)_3 = \\ = \left(\frac{3}{-2 + (z - 1)\omega}\right)_3 = \omega^{1/3(z - 1)} \ne 1 \\ \end{split} \end{equation*} Hence, $z + 1 + 2\omega$ has prime divisor $\pi$ such that 3 is a cubic nonresidue modulo $\pi$ and $\nu_{\pi} (z + 1 + 2\omega)$ is not divisible by 3. The same applies to complex conjugate $\overline{\pi} \mid z - 1 - 2\omega$. Thus, we can take $q = \pi \overline{\pi} \in \mathbb{Z} $.