Solution:

Obviously, for the tables $1 \times n$ and $n \times 1$, the largest number of black cells is $n$. Therefore, we assume that $m \geq 2$ and $n \geq 2$ for the rest of the solution. In the table $m \times n$, we can color the first two rows, the first column and the last column, which is $2 m + 2 n - 4$ black cells in total. It is easy to see that such a table contains no bishop circuit.

Now we show that if there is no bishop circuit, there are at most $2 m + 2 n - 4$ black cells in the table $m \times n$. We denote the cell in $i$-th row and $j$-th column by $(i, j)$. The $k$-th positive diagonal is a set of cells $(i, j)$, such that $i + j - 1 = k$. Similarly, the $k$-th negative diagonal is a set of cells $(i, j)$, such that $n + i - j = k$.

Consider a bipartite graph $G$ with partitions
$$A = \left\{a_{1}, a_{2}, \ldots, a_{m+n-1}\right\} \quad \text{and} \quad B = \left\{b_{1}, b_{2}, \ldots, b_{m+n-1}\right\}$$
where the vertices $a_{i}$ and $b_{j}$ are connected by an edge if and only if the cell in the intersection of the $i$-th positive diagonal and the $j$-th negative diagonal is black. Notice that a bishop circuit corresponds to a circuit in $G$ and vice versa.

The graph $G$ has at least two components: If we color the cells of the table alternately green and red like in chess, then the edges of $G$ corresponding to green cells lie in a different component than the edges of $G$ corresponding to red cells - it is not possible to move a bishop between a green and a red cell.

Furthermore, $G$ has $2 n + 2 m - 2$ vertices. If $G$ is acyclic, then $G$ is a forest consisting of at least two trees. Therefore, $G$ contains at most $2 n + 2 m - 2 - 2 = 2 n + 2 m - 4$ edges and that is the upper bound on the number of black cells we wanted to prove.

Solution:

Denote by $S$ the coloring configuration from the first proof consisting of the first two rows and the first and the last column of the table. It is easy to see that $S$ does not contain the bishop circuit. Also, $S$ is maximal in the sense that if we add any new cell to it, the new coloring will contain a bishop circuit.

We will show that any optimal coloring $C$ has the same number of cells as $S$ by transforming $C$ to $S$ by iterating the following steps:

1) First we choose any cell $a$ which is in $S$, but not in $C$. If there is no such cell, we are done since maximality of $S$ and optimality of $C$ imply that $S = C$.

2) From the optimality of $C$ it follows that there is a bishop circuit $B$ in the coloring $C \cup\{a\}$ containing $a$. Since $S$ does not contain a bishop circuit, there is an element $b$ in cycle $B$ which is not in $S$. We replace coloring $C$ with the coloring $\tilde{C} = (C \cup\{a\}) \backslash\{b\}$.

To finish the proof, we need to show that $\tilde{C}$ is optimal. For that we need to prove that $B$ is a unique cycle in $C \cup\{a\}$ containing $a$.

Assume the opposite. Let $a_{0}, a, a_{1}, \ldots, a_{2 r}$ and $b_{0}, a, b_{1}, \ldots, b_{2 s}$ be two bishop cycles in $C \cup\{a\}$ such that $a_{0}$ and $b_{1}$ (as well as $a_{1}$ and $b_{0}$) are on the same diagonal. Consider the cycle in $C$ (every two consecutive cells are on the same diagonal)
$$a_{0}, b_{1}, b_{2}, \ldots, b_{2 s}, b_{0}, a_{1}, a_{2}, \ldots, a_{2 r}$$
It remains to prove that it contains a bishop circuit which will contradict the optimality of $C$.

Note that no three consecutive cells are on the same diagonal, so the only problem is if the cells are not pairwise different. Thus we can assume that we can write the cycle in the following form
$$c_{1}, \ldots, c_{k}, c_{1}, d_{1}, d_{2}, \ldots, d_{t-1}$$
where $t \geq k$. If we remove first $k$ cells we obtain the cycle
$$c_{1}, d_{1}, d_{2}, \ldots, d_{t-1}$$
Furthermore, if $d_{t-1}, c_{1}$ and $d_{1}$ are on the same diagonal, we remove $c_{1}$. By repeating this procedure, we end up with the bishop's circuit.