About a field of order $2^{n}$ with $n$ an odd integer and an additional property I'm new in the world of fields (so I don't have any strong theorem at my disposal) and I've got stuck in this problem:

Given a field of order $2^{n}$ with $n$ an odd integer and $a,b$
  elements in the field such that $a^2+ab+b^2=0$, prove that $a=b=0$.

I've found this answer  but in the book i'm studying from (Gallian's book) we've not touched that stuff, the only thing I can use is that the characteristic is two and the fact that this fields forms a group under multiplication. I'm looking for something like this in the general case.
Please, if someone can help me that would be great.

The proof that introduces degrees, irreducible polynomials, etc. can be "written down" to an explicit calculation.
If either $a,b$ is zero, then $a^2 + ab + b^2 = 0$ implies both are.  So assume neither $a,b$ is zero, and in particular $x = a/b$ is nonzero.
As before $x^2 + x + 1 = 0$, or as we are in characteristic 2, $x^2 = x+1$ and $x(x+1) = 1$.  That is, in the multiplicative group of the field, $x$ has order 3:
$$ x^3 = x(x+1) = 1 $$
Therefore 3 must divide the order of the multiplicative group $2^n-1$.  But this is only possible when $n$ is even by a simple congruence calculation, $2^2 \equiv 1 \pmod{3}$.
