Here's a naive solution that uses only the definition of linear independence, the notion of dimension, and the fact that $\dim \Bbb F^2 = 2$.
One direction is trivial. For the other direction, for clarity of presentation we'll give an explicit argument for $k = 3$, but the general case is totally analogous:
Suppose $A^2 \neq 0$; then, there is a vector $\xi \in \Bbb F^2$ such that $A^2 \xi \neq 0$. We'll show that the triple $(\xi, A \xi, A^2 \xi)$ is linearly independent, which implies that $\dim \Bbb F^2 \geq 3$, a contradiction:
Suppose not; then, there are coefficients $a, b, c$, not all $0$, such that $$0 = a \xi + b A \xi + c A^2 \xi.$$ Multiplying both sides of this equation by $A^2$ gives
$$0 = a A^2 \xi + b A^3 \xi + c A^4 \xi = a A^2 \xi$$ (the second equality uses that $A^3 = 0$). Since $A^2 \xi \neq 0$, we must have $a = 0$ and so $$0 = b A \xi + c A^2 \xi .$$ Multiplying both sides by $A$ similarly leads to $b = 0$, leaving $$c A^2 \xi = 0,$$ so $c = 0$, a contradiction.