When $A$ is $3\times3$, prove that $A^2=0$ iff $A$ has rank less than or equal one and trace zero

Suppose that $A^2 = 0$; we must have $\operatorname{Im}(A) \subset \ker(A)$. By the rank-nullity theorem, we can therefore conclude that the rank of $A$ (the dimension of $\operatorname{Im}(A)$) is at most $1$. Now, it suffices to use the fact that $A^2 = \operatorname{trace}(A)\cdot A$, as you noted.