My only idea on how to approach this problem was by making the matrix upper triangular. However, I found a simple counterexample to fact that elementary transitions do not change nilpotence of matrix. It does. If it was illegal, I would just say that it is nilpotent if $a_{11} \ne 0$.
Do you have any better ideas on how to solve this problem? Thank you in advance.
A rank one matrix can be written as $A = v \cdot w^T$ where $v \neq 0$ is a non-zero column vector (which spans the range of $A$) and $w$ is a non-zero row vector. If you expand the product $A = v \cdot w^T$ you will see that the columns of $A$ are precisely $w_1 v, \dots, w_n v$. Now, note that
$$ A^2 = v \cdot w^T \cdot v \cdot w^T = v \cdot \underbrace{\left( w^T \cdot v \right)}_{1 \times 1 \textrm{ matrix, a scalar}} \cdot w^T = (w^T \cdot v) \cdot v \cdot w^T = cA $$
where $c = w^T \cdot v$. By induction you get that $A^k = c^{k-1} A$ and so $A$ is nilpotent if and only if $c = 0$ (in which case, $A^2 = 0$ so that $A$ is nilpotent of nilpotency rank two).
To interpret $c = w^T \cdot v = \sum_{i=1}^n w_i v_i$ note that $c$ is precisely the trace of $A$ so that $A$ is nilpotent if and only if the sum of the diagonal elements of $A$ is zero.