As many ones as possible, square zero

Idea. With nonnegative entries, $A^2=0$ means no index can both receive and send a one. So the ones live in a block whose rows and columns use disjoint sets of indices, and such a block has at most $3\times 3=9$ cells.

Step 1: no cancellation. $(A^2{)}_{ik}={\sum }_j{a}_{ij}{a}_{jk}$ is a sum of nonnegative terms, so it is zero only if $a_{ij}{a}_{jk}=0$ for every $j$. Hence there is no $j$ with a one in column $j$ (some $a_{ij}=1$) and a one in row $j$ (some $a_{jk}=1$).

Step 2: the bound. Let $S$ be the set of rows that contain a one and $T$ the set of columns that contain a one. By Step 1, $S$ and $T$ are disjoint subsets of $\{1,\dots ,6\}$. Every one lies in the block $S\times T$, so the number of ones is at most $$|S|\cdot |T|\le |S|\cdot (6-|S|)\le 3\cdot 3=9.$$

Step 3: a matrix with 9 ones. Put ones in rows 1 to 3 and columns 4 to 6, and zeros elsewhere: $$A=\left(\begin{array}{cc}0& J\\ 0& 0\end{array}\right),$$ with $J$ the $3\times 3$ block of ones. Then $A^2=0$ because every one points from $\{1,2,3\}$ to $\{4,5,6\}$, and nothing leaves $\{4,5,6\}$.

So the answer is 9. For an $n\times n$ matrix it is $\lfloor n^2/4\rfloor$.

Remark. The nonnegativity is essential. Over the real numbers, without the 0-1 condition, $A^2=0$ allows every entry to be nonzero, for example $A=\left(\begin{array}{cc}1& -1\\ 1& -1\end{array}\right)$.