Suppose that $a$ and $b$ belong to a field of order $8$ and $a^2 + ab + b^2 =0$. Then $a=0$ and $b=0$. Do the same when the field has order $2^n$ with $n$ odd?
If one of the term is zero, i.e. let $b=0$ then $a^2 =0 \implies a =0$. We also note that since field is of order $8$, $a^7 = 1 = b^7$.
But how to bring a contradiction if I consider any one to be not equal to $0$? I want to use only basics.
On
First we have
$(a-b)(a^2+ab+b^2)=a^3-b^3=0$, or $a^3=b^3$.
Since $a^7=b^7=1,$
$a^7=a^4a^3=a^4b^3=1$, $b^7=b^3b^4=a^3b^4=1$, and $a^4b^3=a^3b^4$.
So $a^3b^3(a-b)=0$.
If $a^3=0$, then $a=0$, and then $b^2=0$, so $b=0$. Similarly it is true for $b^3=0$.
Finally, if $a^3\neq0$ and $b^3\neq0$, then $a=b$. So there is $3a^2=0$, and thus
$a=0$ and $b=0$.
If one of them, say $b$, is not $0$, then $\frac{a}{b}$ is a root of $x^2+x+1$, which lies in a (the) degree-$2$ extension of $\mathbb F_2$, and therefore it does not lie in the degree-$3$ extension of order $8$. Same is true for any odd $n$ and for the same reason.