Zeta-function of $x_0^3 + x_1^3 + x_2^3 = 0$ over $\mathbb{F}_4$.

Using some facts about elliptic curves it could go as follows.

Let's look at the cubic projective curve given by the equation $$ X^3+Y^3+Z^3=0. $$ Putting this into Weierstrass form is easy. All we need to do is to substitute $Z=Y+U$, and (recalling that we are in characteristic two) we get $$ X^3+U^3+U^2Y+UY^2=0. $$ If we dehomogenize $x=X/U,y=Y/U$, we get an equation in the Weierstrass form $$ E:x^3+1=y^2+y.\qquad(*) $$ The above process maps $\Bbb{F}_{4^m}$-rational points to each other bijectively, so we might as well count the number of points $\#E(\Bbb{F}_{4^m})$.

The curve $E$ has a very special form. If $P=(x_0,y_0)\in E$, then the negative of this point (= the other point on the same vertical line) is $[-1]P=(x_0,y_0+1)$. Furthermore, implicit differentiation tells us that the tangent of $E$ at $P$ has slope $x_0^2$. The usual process then leads to the very simple point doubling formula $$ [2](x_0,y_0)=(x_0^4,y_0^4+1).\qquad(**) $$ In other words doubling can be achieved by applying the square of the Frobenius followed by negation.

If $(x_0,y_0)\in E(\Bbb{F}_{4^m})$ then repeated application of $(**)$ tells us that $$ [2^m](x_0,y_0)=(x_0^{4^m},y_0^{4^m}+m\cdot1)= \begin{cases}P,&\ \text{if $m$ is even, and}\\ [-1]P,&\ \text{if $m$ is odd.} \end{cases} $$

This means that all the points $P$ of $E(\Bbb{F}_{4^m})$ satisfy the equation $[M]P=0$ for $M=2^m-(-1)^m$.

With a few basic facts about elliptic curves in place this implies that $$E(\Bbb{F}_{4^m})=E[M],\qquad(**)$$ and thus has $M^2$ points. More precisely, the highlighted result tells that we have an inclusion "$\subseteq$" in $(**)$, so $\#E(\Bbb{F}_{4^m})$ must be a factor of $M^2$. But then the Hasse-Weil bound $|\#E(\Bbb{F}_{4^m})-(4^m+1)|\le2\cdot 2^m$ rules out all proper factors.

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I'm afraid this is unsatisfactory as an answer here. Hasse-Weil bound was needed to conclude, and that is at the same depth as the use of Zeta functions. I posted it chiefly, because I discussed an argument related to elliptic curve in the comments. The calculation I recalled must have been about another elliptic curve defined over $\Bbb{F}_2$. Sorry. Anyway, this does lead to the conclusion, but overall the method is kludgier than those listed in the question.