Placing checkers on an m x n board 
What is the maximum amount of checkers you can place on an $m \times n$ checkerboard such that no four checkers make a rectangle parallel to the rows and columns?

The square colors don't matter; it is essentially a grid. Also the checkers are all indistinguishable.
I tried looking at the problem row by row, but this devolved into too many cases. I've also looked at small values of $m$, but couldn't get anywhere.
It seems that if there are a lot of rows, the optimal way is to place 2 checkers for the first $\binom{n}{2}$ rows, and then place 1 continuing down the rest. However, when there are relatively few rows, there seems to be a lot of cases regarding how many checkers to place for each row.
 A: Some simple observations
Let $\mathscr{C}(m,n)$ be the maximum number of checkers one can place on an $m\times n$ board without any four of them forming a rectangle with sides parallel to the board.  We will refer to $m$ as the number of rows and $n$ as the number of columns.
The roles of $m,n$ are interchangeable, so the function $\mathscr{C}(m,n)$ is symmetric in the sense that $\mathscr{C}(m,n)=\mathscr{C}(n,m)$.
The function is strictly increasing with either argument:
$$ \mathscr{C}(m+1,n) \ge \mathscr{C}(m,n) + 1$$
and when $m,n \gt 1$ we have:
$$ \mathscr{C}(m+1,n+1) \ge \mathscr{C}(m,n)+ 3 $$
For smallish values of $m$ we can work out the maximum number of checkers:
$$ \mathscr{C}(1,n) = n \; ; \; \mathscr{C}(2,n) = n+1 $$
and if $m \ge \binom{n}{2}$ the exact maximum can be reached:
$$ \mathscr{C}(m,n) = \binom{n}{2} + m $$
by placing two checkers on each of $\binom{n}{2}$ rows and one checker on each of the remaining rows (if any).
The challenge is then to obtain good estimates for the intermediate cases, especially when $m,n$ are approximately equal.
Partial pairwise block designs
Consider an $m\times n$ incidence matrix with $1$'s in place of checkers on a grid, otherwise $0$'s.  In this way we define a combinatorial design, not necessarily uniform, with the rows representing $m$ blocks (lines) and the columns representing $n$ varieties (points).
The requirement that no rectangle is formed by the checkers amounts simply to the condition that no pair of column indexes $(j_1,j_2)$ appears in more than one block (row), equivalently no pair of rows (lines) intersect in more than one point.  We do not require that all blocks have an equal number of points (which defines a uniform block design) nor that all points are incident with an equal number of lines (a regular block design).  Some may find the language of finite geometries more appropriate here.
When any two distinct points lie in a unique common block, we have a pairwise balanced design (PBD), for which there are many known results.   A famous result by R. Wilson in the early seventies says such a design exists, when all the blocks are of equal size $k$, for all sufficiently large $n$ provided there are divisibility conditions:
$$ (k-1) \mid (n-1) \;\text{ and }\; k(k-1) \mid n(n-1) $$
Often finite projective planes afford especially large such designs, although these are only known to exist for $m=n=q^2+q+1$ for a prime power $q$ (it is an open problem whether projective planes of other sizes exist).
When $q=2$ we have the Fano plane, which would give us a $7\times 7$ board with three checkers in each row (resp. in each column):
$$ \begin{bmatrix} 1 & 1 & 1 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 1 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 & 1 & 0 \end{bmatrix} $$
More generally these block designs will have $q+1$ points in each block (and dually each point belongs to $q+1$ blocks), so there is no fixed upper bound on the number of checkers that can share a row or column of the board.
Focusing on the standard $8\times 8$ board, we easily add three more checkers (points) to the $7\times 7$ design above, using an extra row and column.  This gives $24$ checkers on a standard board:
$$ \begin{bmatrix} 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\
0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 \\
0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 1 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} $$  
Further case analysis, easy but tedious, proves this is a maximum.  (Caveat: I did this analysis by hand, and I plan to follow up with code to double check and extend it.)
A: A longish comment:
The number of possible rectangles is given by the number of ways to choose two columns times the number of ways to choose two rows. i.e.:
$$\frac{nm(n-1)(m-1)}{4}$$
The number of ways to put $4$ checkers on the board is:
$${nm}\choose{4}$$
Thus, the number of ways to put $4$ checkers on a board without rectangles is just:
$${{nm}\choose{4}} - \frac{nm(n-1)(m-1)}{4}$$
This does not answer your question regarding $k$ checkers on the board
