Suppose $A$ is a complex nilpotent $2\times 2$ matrix, $AB = - BA$. Prove that either $A = 0$ or $\operatorname{tr}(B) = 0$.
Since $\operatorname{tr}(XY) = \operatorname{tr}(YX)$ for arbitrary matrices $X, Y$ and $AB = -BA$: $$\operatorname{tr}(AB) = \operatorname{tr}(BA) = - \operatorname{tr}(BA) \Rightarrow \operatorname{tr}(BA) = \operatorname{tr}(AB) = 0.$$
The trace of nilpotent matrix is zero, so $\operatorname{tr}(A) = 0$.
If $A = 0$, then everything is correct. I can't finish the proof and deduce obstruction from the case $A \neq 0$ and $\operatorname{tr}(B) \neq 0$.
You can always write a $2\times 2$ nilpotent matrix in the form $$A=\begin{bmatrix}a&b\\c&-a\end{bmatrix}:bc=-a^2$$ since $\operatorname{tr}(A)=0$ and $\det(A)=0$ (indeed, see the second answer here). So suppose $A\neq 0$. Then set $$B:=\begin{bmatrix}w&x\\y&z\end{bmatrix}$$ and use $AB=-BA$ to compare entries and show that $w=-z$. Notice that there are only two cases: $(1)$ is that $c\neq 0$, or $(2)$ is that $c=0$ and in this case we must have $b\neq 0$ (else, $c=b=0$ implies $a=0$ and thus $A=0$, contradicting our supposition).