The conjecture is true. To prove it, we can restrict to $k=1$ as explained in the comments. Let $\chi$ denote a cubic Dirichlet character modulo $p$; this exists as $p\equiv 1\pmod{3}$, and it is unique up to complex conjugation. Let $\psi$ denote the (unique) quadratic Dirichlet character modulo $p$.

The number of points of the *affine* elliptic curve $E$ modulo $p$ equals
$$\sum_{y\bmod p}\left(1+\chi(y^2-2)+\overline{\chi}(y^2-2)\right)=p+\sum_{y\bmod p}\chi(y^2-2)+\overline{\sum_{y\bmod p}\chi(y^2-2)}.$$
The $y$-sum on the right hand side equals
$$\sum_{y\bmod p}\chi(y^2-2)=\sum_{z\bmod p}\chi(z-2)\bigl(1+\psi(z)\bigr)=\sum_{z\bmod p}\chi(z-2)\psi(z).$$
Here $\chi(z-2)=\chi(2-z)$, because $\chi(-1)=\chi((-1)^3)=\chi^3(-1)=1$. Therefore,
$$\sum_{y\bmod p}\chi(y^2-2)=\sum_{z\bmod p}\chi(2-z)\psi(z)=\chi(2)\psi(2)J(\chi,\psi),$$
where $J(\chi,\psi)$ is the corresponding Jacobi sum.

To evaluate the Jacobi sum $J(\chi,\psi)$, we rely on Chapter 6 of Rose: A course in number theory (2nd ed., Oxford University Press, 1994). By (the last part of) Exercise 10 at the end of this chapter,
$$J(\chi,\psi)=\chi(4)J(\chi,\chi).$$
Now we observe that $4p=1+27(2a+1)^2$, and this is the only way to write $4p$ as $u^2+27v^2$ with $u\equiv 1\pmod{3}$ and $v$ positive, cf. Theorem 2.5 and the subsequent comments. Therefore, by Lemma 2.6,
$$J(\chi,\chi)=(3a+2)+(6a+3)e^{\pm 2\pi i/3}=\frac{1\pm(6a+3)i\sqrt{3}}{2},$$
the choice of the $\pm$ sign depending on which cubic character we denoted by $\chi$.

Putting everything together,
$$\sum_{y\bmod p}\chi(y^2-2)=\chi(2)\psi(2)J(\chi,\psi)=\chi(8)\psi(2)\frac{1\pm(6a+3)i\sqrt{3}}{2}.$$
Here, $\chi(8)=\chi(2^3)=\chi^3(2)=1$, and so
$$\sum_{y\bmod p}\chi(y^2-2)+\overline{\sum_{y\bmod p}\chi(y^2-2)}=\psi(2)\sum_{\pm}\frac{1\pm(6a+3)i\sqrt{3}}{2}=\psi(2).$$

To summarize, the number of points of the *affine* elliptic curve $E$ modulo $p$ equals $p+\psi(2)$.

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