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.

Or you could do it entirely with group theory: Suppose that $a \neq b$ .  Notice that $a^{3}-b^{3} = (a-b)(a^{2}+ab+b^{2})=0,$ so $a^{3} = b^{3}.$ We claim that $m = 2^{n }-1$ must be divisible by $3.$ This is the order of the order of the multiplicative group of the field containing $a$ and $b,$ so we have $a^{m} = b^{m} = 1.$ Now if $3$ does not divide $m,$ then we have $3x + my = 1$ for integers $x$ and $y.$ Then $a = a^{3x+my} = a^{3x} = b^{3x} = b^{3x+my} = b^{1} = b,$ a contradiction. Hence $3$ divides $2^{n}-1,$ so as noted by others $n$ must be even.
