Let $A$ be a binary $n \times n$ matrix, such that $A^2=0$. What is the max num of $1$'s that $A$ could have?

I know that you ask about non-graph approaches in particular. I still felt like a complete proof using graph theory might add something to this thread. I use exactly the technique that you propose in your post.

We want to prove that the maximum number of $1$'s in an $n \times n$ binary matrix $A$ with $A^2 = 0$ is $\lfloor \frac{n}{2} \rfloor \lceil \frac{n}{2} \rceil$.

Let $A$ be an arbitrary matrix of this form. Consider the directed graph $G$ with adjacency matrix $A$. As pointed out in the question, $A^2 = 0$ implies that the longest directed path in $G$ has length at most $1$. Hence no vertex in $G$ has an outgoing and an incoming edge. Therefore there exist only sources, sinks and isolated vertices in $G$. Let $U$ be the set of vertices which are sources and let $V$ be the set of vertices which are sinks or isolated vertices. Clearly all edges in $G$ must go from $U$ to $V$. Hence there are at most $|U||V| = |U| (n - |U|)$ edges in $G$.

The function $f(k) = k(n - k) = nk -k^2$ has a global maximum at $k = \frac{n}{2}$. The two global integer maxima are at $k = \lfloor \frac{n}{2} \rfloor$ and $k = \lceil \frac{n}{2} \rceil$ with value $f(\lfloor \frac{n}{2} \rfloor) = f(\lceil \frac{n}{2} \rceil) = \lceil \frac{n}{2} \rceil \lfloor \frac{n}{2} \rfloor$. Applying this to our argument above we get that there are at most $\lceil \frac{n}{2} \rceil \lfloor \frac{n}{2} \rfloor$ edges in $G$.

Every edge in $G$ corresponds to exactly one $1$-entry in $A$. Hence $A$ has at most $\lceil \frac{n}{2} \rceil \lfloor \frac{n}{2} \rfloor$ entries with a $1$.

We still have to prove that there actually exists a binary $n \times n$ matrix $A$ with $A^2 = 0$ and with exactly $\lceil \frac{n}{2} \rceil \lfloor \frac{n}{2} \rfloor$ $1$-entries. Such a matrix can be constructed as follows:

$$ A_{i, j} = 1 \iff i \leq \frac{n}{2} \; \And \; j > \frac{n}{2}$$

Let $R=\{i | \text{ row i is not empty}\}$ and let $C= \{ i | \text{ column i is not empty}\}$. Notice $A^2=0$ is equivalent to $R\cap C = \varnothing$. Hence the maximum is achieved when $a_{i,j} = 1 \iff i\in R, j\in C$. So we want to maximize $|R||C|$ which clearly happens when $|R|$ and $|C|$ sum to $n$ and differ by at most $1$.