A property of finite field of order $2^n$

If $b$ were not 0 then $a/b$ would be a root of $x^2 + x + 1$, which is irreducible over ${\mathbf F}_2$. Look at the size of the field ${\mathbf F}_2(a/b)$ and the size of the field you are working in that has order $2^n$.

If $a=b$, then $a^2+ab+b^2=3a^2=a^2=0$ and $a=0=b$, we are done.

Suppose that $a\neq b$.

Observe that $0=(a-b)(a^2+ab+b^2)=a^3-b^3$. Thus, $a^3=b^3$.

We claim that $a=0$ and $b=0$.

If $a\neq 0$, then $(a^{-1}b)^3=1$ and the multiplicative order of $a^{-1}b$ in the multiplicative group $F-\{0\}$ is $1$ because $3\not\mid |F-\{0\}|=2^n-1$. Hence, $a^{-1}b=1$ and $a=b$, a contradiction. Therefore, $a=0$. By a similar argument, we have $b=0$.