We have three possible cases.

The first case is when there are no two adjacent cells with the same color. In this case there are exactly two possible ways to color this $n \times n$ chessboard depending on the color of the first cell on the top left of the chessboard:

The second case is when there are two adjacent cells with the same color on the same row.

In this case the two adjacent cells in the previous row and the two adjacent cells in the next row must be of the other color.

If the previous cell or the next cell in this row is of the same color, the adjacent cells in the previous or the next row must be of the other color. If the previous cell or the next cell in this row is of the other color, the adjacent cells in the previous or the next row must be of this color.

Therefore, once fixing the colors of the cells of this row, the colors of the cells of the other rows are fixed by just alternating vertically the colors of adjacent cells.

Hence, in this case any coloring of the first row with at least two adjacent cells with the same color will fix the coloring of the cells of the whole chessboard. There are precisely $2^{n}-2$ possible such coloring corresponding to the number of subsets of $\{1,2, \ldots, n\}$ (the positions of the black cells in the row) minus the two alternating coloring of the first case.

The third case is when there are two adjacent cells with the same color on the same column. This case is similar to the previous one and in the same way we obtain $2^{n}-2$ possible such coloring.

Notice that the second and the third cases cannot occur at the same time. Therefore, there are precisely $2+2\left(2^{n}-2\right)=2^{n+1}-2$ possible coloring of the $n \times n$ chessboard.