Edit. In hindsight, the original proof is overly complicated. It can be made somewhat simpler:
As $A^2+B^2=I$, the matrix square $A^2$ commutes with all polynomials in $B$. Hence $A$ commutes with $C=AB-BA$. Now, if $C$ is nonsingular, then $C^{-1}$ commutes with $A$ too and in turn
$$
I=(AB-BA)C^{-1}=ABC^{-1}-BAC^{-1}=ABC^{-1}-BC^{-1}A.
$$
Taking traces on both sides, we get $n=0$, which is a contradiction because $n$ is odd. $\square$
(The original proof goes here.)
The techniques for proving the classical result that "$A$ commutes with $[A,B]$ implies that $[A,B]$ is nilpotent" (over a field of characteristic zero) can be reused here.
First, as $A^2+B^2=I$, the matrix square $A^2$ commutes with all polynomials in $B$. Therefore $A$ commutes with $AB-BA$ when $1=-1$.
Next, if we define an iterate $X_1 = B$ and $X_{k+1}=X_k(AB-BA)$ for every $k\ge1$, one can prove by mathematical induction that $(AB-BA)^{k+1}=AX_k-X_kA$ and hence all positive powers of $C=AB-BA$ are traceless.
Now, let $\lambda_1,\ldots,\lambda_r$ be those distinct and nonzero eigenvalues of $C$ over the algebraically closure of the underlying field and $m_1,\ldots,m_r$ be their algebraic multiplicities respectively, so that $m_1+\ldots+m_r\le n$ and equality holds iff $C$ is nonsingular. Then the condition $\operatorname{tr}(C^k)=0$ for every $k\ge1$ results in a Vandermonde system of equations:
$$
\pmatrix{
\lambda_1&\lambda_2&\cdots&\lambda_r\\
\lambda_1^2&\lambda_2^2&\cdots&\lambda_r^2\\
\vdots&\vdots&&\vdots\\
\lambda_1^r&\lambda_2^r&\cdots&\lambda_r^r\\
}
\pmatrix{m_1\\ m_2\\ \vdots\\ m_r}=0.
$$
If this were a field of characteristic zero, the above would imply that all $m_i$s are zero and $C$ is nilpotent. However, as we are working over a field of characteristic $2$ here, the above only implies that every $m_i$ is congruent to zero modulo $2$, i.e. every $m_i$ is even. Thus $m_1+\ldots+m_r<n$ because $n$ is odd, meaning that $C$ has a zero eigenvalue.
