Consider $F := GF(q)$ where $q = p^e$ and $E := GF(q^2)$ where w is a primitive element of $E$.
Fix $\theta := w^{q - 2}$.

Starting point: can I always write $1 + \theta$ as a power of $\theta$?

If $Gcd (q^2 - 1, q - 2) = 1$, $\theta$ is a primitive element for $E$ and the answer is yes.

So we can restrict to the case $q \mod 3 = 2$ where all cubes of elements of $E$ are powers of $\theta$.

Is $1 + \theta$ a cube in $E$? No, when $e$ is odd; otherwise yes.

So in the remaining odd case we want to replace the role of $1 + \theta$ by $1 + \theta^j$ for some $j$, leading to the following conjecture.

Conjecture: Consider $q = p^e$ where $p = 2 \mod 3$ and $e$ is odd. There exists $j$ in $\{1, \dots, q^2 - 1\}$ such that $Gcd (q^2 - 1, j) = 1$ and $1 + \theta^j$ is a cube in $E$.

Computational evidence in support of this conjecture is strong. While it does not hold for $q = 5$ and $q = 8$, such $j$ exists for all relevant $q$ in the range $[11\dots 10^8]$. Always $j$ is at most $q - 1$.

The context for the query is the study of presentations for the classical groups $SU(3, q) \leq GL(d, q^2)$ where a generator has action described by $\theta$.

Pointers towards a proof would be much appreciated.