Let the columns of $A$ and $B$ be $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ respectively. By definition, the rank of $A$ and $B$ are the dimensions of the linear spans $\langle a_1, \ldots, a_n\rangle$ and $\langle b_1, \ldots, b_n\rangle$. Now the rank of $A + B$ is the dimension of the linear span of the columns of $A + B$, i.e. the dimension of the linear span $\langle a_1 + b_1, \ldots, a_n + b_n\rangle$. Since $\langle a_1 + b_1, \ldots, a_n + b_n\rangle \subseteq \langle a_1, \ldots, a_n, b_1, \ldots, b_n\rangle$ the result follows.
Edit: Let me elaborate on the last statement. Any vector $v$ in $\langle a_1 + b_1, \ldots, a_n + b_n\rangle$ can be written as some linear combination $v = \lambda_1 (a_1 + b_1) + \ldots + \lambda_n (a_n + b_n)$ for some scalars $\lambda_i$. But then we can also write $v = \lambda_1 (a_1) + \ldots + \lambda_n (a_n) + \lambda_1 (b_1) + \ldots + \lambda_n (b_n)$. This implies that also $v \in \langle a_1, \ldots, a_n, b_1, \ldots, b_n\rangle$. We can do this for any vector $v$, so
$$\forall v \in \langle a_1 + b_1, \ldots, a_n + b_n\rangle: v \in \langle a_1, \ldots, a_n, b_1, \ldots, b_n\rangle$$
This is equivalent to saying $\langle a_1 + b_1, \ldots, a_n + b_n\rangle \subseteq \langle a_1, \ldots, a_n, b_1, \ldots, b_n\rangle$.
That follows from the book Simultaneous Triangularization by Radjavi and Rosenthal (page 8). The original proof is due to Thomas Laffey.
Let $\{y\}$ be a basis of $\mathrm{Im}(AB-BA)$. Let $\lambda\in\mathrm{Spec}(B)$. If $B=\lambda I$, then there is almost nothing to do. Otherwise $F=\ker(B-\lambda I)$, $G=\mathrm{Im}(B-\lambda I)$ are non-trivial $B$-invariant subspaces. If we show that $F$ or $G$ is $A$-invariant, then we are the kings of oil.
Assume that $F$ is not $A$-invariant. Then there is $x$ s.t. $(B-\lambda I)x=0$, $(B-\lambda I)Ax\not= 0$. We have $$A(B-\lambda I)x-(B-\lambda I)Ax=ABx-BAx=-(B-\lambda I)Ax\in\mathrm{Im}(AB-BA)\cap\mathrm{Im}(B-\lambda I)\setminus\{0\}.$$ Thus $y\in G$.
Let $z\in \mathbb{C}^n$. Then $A(B-\lambda I)z$ is in the form $(B-\lambda I)Az+\alpha y$. Therefore, $G$ is $A$-invariant and we are done. $\square$
Best Answer
Since $\text{rank}(AB-BA) = 1$, we can write $AB-BA = uv^*$ for some vectors $u,v \in \mathbb{C}^n$.
Since $\text{tr}(AB) = \text{tr}(BA)$, we have $0 = \text{tr}(AB-BA) = \text{tr}(uv^*) = v^*u$.
Hence, $(AB-BA)^2 = (uv^*)^2 = u\underbrace{v^*u}_{= 0}v^* = 0_{n \times n}$, i.e. $AB-BA$ is nilpotent.