For rook we have obviously $$R(n,m)=nm(n+m-2)$$ and for bishop $$B(n,m)=4\left(m\binom{n}{2}-\binom{n+1}{3}+\binom{n-m+1}{3}\right)$$ if we assume $\binom{n}{k}=0$ for $n<0$.
Is there a way to write it in more simple and nice form?
For rook we have obviously $$R(n,m)=nm(n+m-2)$$ and for bishop $$B(n,m)=4\left(m\binom{n}{2}-\binom{n+1}{3}+\binom{n-m+1}{3}\right)$$ if we assume $\binom{n}{k}=0$ for $n<0$.
Is there a way to write it in more simple and nice form?
As greedoid commented, it is so simple to expand and simplify. $$B(n,m)=4\left(m\binom{n}{2}-\binom{n+1}{3}+\binom{n-m+1}{3}\right)=\frac{2}{3} m(m-1) (3 n-m-1)$$
The number of possible bishop moves on an $m\times n$ chessboard is equal to $4$ times the number of $k\times k$ squares, $2\le k\le\min(m,n)$, on that chessboard, since each bishop move is a move from one corner to the opposite corner of one of those squares.
From this we get the formula $$B(m,n)=4\sum_{k=2}^{\min(m,n)}(m+1-k)(n+1-k)$$$$=8\binom{\min(m,n)}3+4(|m-n|+1)\binom{\min(m,n)}2$$ which you may or may not consider "simple and nice". For square chessboards it becomes simpler and nicer: $$B(n,n)=4\sum_{k=2}^n(n+1-k)^2=4\left[1^2+2^2+3^2+\cdots+(n-1)^2\right]$$$$=8\binom n3+4\binom n2=\frac23(n-1)n(2n-1).$$
See also the answer to the question How many ways can you put: a) two bishops b) two knights c) two queens on a chessboard in such a way that one piece does not attack the other?.