Cute for general p > 17, take x = (b-a) and y = (c + a) for constant b and c which satisfy 3(b+c) + 2023 is divisible by p. Then we get that x^3 + 2023xy + y^3 is linear in a.

Official Solution: $\textbf{Comment 1.}$ The Lemma is essentially bringing $\textit{Cardano's Formula}$ from $\mathbb{C}$ into $\mathbb{F}_p$: \begin{align*} y^3 + ay +b = 0 \ \Longrightarrow\ y=\alpha_+ +\alpha_- \textit{ where } \alpha_\pm^3 = \frac{-b \pm \sqrt{b^2 + \frac{4}{27}a^3}}{2} \textit{ and } \displaystyle \alpha_+\alpha_- = -\frac{a}{3}. \end{align*}The last condition is required because in $\mathbb{C}$ there are always three cube roots, giving potentially nine choices. Here we just brought the formula into $\mathbb{F}_p$ with $(a, b) = (kx, x^3 - n)$ and carefully handled the square root term. Introducing $\mathbb{F}_p[\delta]$ in Case 2 is in fact similar to how complex numbers were used to deal with cases where Cardano's Formula contains terms with square root of negative numbers, even though the result is a real number. The key is that even though $\delta \not\in \mathbb{F}_p$, the cube root terms eventually cancel out as conjugates. $\textbf{Comment 2.}$ If one recognises $\mathbb{F}_{p}[\delta] \cong \mathbb{F}_{p^2}$ is a cyclic multiplicative group, the Criterion is unnecessary as $(**)$ and $3 \nmid p+1$ directly tells us that $\dfrac{k-x^3 +\delta}{2}$ has a cube root. $\textbf{Comment 3.}$ If we were to generalise the problem by changing 2023 to any integer $N$. The answer will then be \begin{align*} \begin{cases} p\neq 7 & \text{if $7 \mid N$}\\ \forall p & \text{if $7 \nmid N$} \end{cases} \end{align*}