dealing with the more difficult direction - i will leave the details for you to complete, since this is a useful exercise to go through, in terms of familiarizing yourself with use of the index notation, and gaining practice in thinking at the required level of abstraction. but there may be a fairly intuitive and straightforward solution along the following lines:

suppose $n \ge 2$. let $A$ be an $n \times n$ matrix which acts on the $n-$ dimensional vector space $F^n$.

first you may show that if any $2\times 2$ minor of $A$ does not vanish then the image has dimension at least $2$. by permuting rows and columns you can see that this is implied by the case for $n=2$ which is straightforward.

second, knowing that all $2\times 2$ minors of a rank 1 matrix must vanish, you have the condition:
$$
a_{ij}a_{kl}=a_{il}a_{kj}
$$

suppose firstly that none of the entries in $A$ is zero. then you should have no trouble in deriving the desired conclusion.

we now need to show that the problem can, in effect, be reduced to the case already dealt with.

if $A \ne 0$ then we have an element, wlog $a_{11} \ne 0$. you may now show that if $a_{1j}=0$ that we must have $a_{kj}=0$ for all $k$.

the same thing follows for columns if you invoke the fact that the rank of a matrix is equal to the rank of its transpose.