For a n*n square, the answer is 2^n. Rather than white and black, fill the board with zeroes and ones. It’s easy to see that adding two boards módulo 2 gives another good board. Additionally, every good board is determined by the lowest row. So to prove my statement it is enough to prove that every lowest row with only black gives a good arrangement. But after playing around it is evident that it is the case, it creates a rotated rectangle of sorts. For n=5 looks like this: 0000x 000x0 00x00 000x0 00x0x 0x0x0 00x00 0x0x0 x0x0x 0x000 x0x00 0x0x0 x0000 0x000 00x00

in above there's n of these rotated rectangles, and they are linearly independent so they span a set of at least 2^n rectangles (adding means toggling) upper bound is because of there is an injection from grid colorings to bottom row colorings

if 5×7 what is your answer

If you paint the last line randomly, there will be 1 case in each case, so the answer is 2^2024.