Let $a_{n}$ denote the number of ways to color a $2 \times n$ grid subject only to the given constraint, and $b_{n}$ denote the number of ways to color a $2 \times n$ grid subject to the given constraint, but with the added restriction that the first column cannot be colored black-black. Consider the first column of a $2 \times n$ grid that is not subject to the additional constraint. It can be colored black-white or white-black, in which case the leftmost $2 \times 2$ square is guaranteed not to be monochromatic, and so the remaining $2 \times(n-1)$ subgrid can be colored in $a_{n-1}$ ways. Otherwise, it is colored white-white or black-black; WLOG, assume that it's colored black-black. Then the remaining $2 \times(n-1)$ subgrid is subject to both constraints, so there are $b_{n-1}$ ways to color the remaining subgrid. Hence $a_{n}=2 a_{n-1}+2 b_{n-1}$. Now consider the first column of a $2 \times n$ grid that is subject to the additional constraint. The first column cannot be colored black-black, and if it is colored white-black or black-white, there are $a_{n-1}$ ways to color the remaining subgrid by similar logic to the previous case. If it is colored white-white, then there are $b_{n-1}$ ways to color the remaining subgrid, again by similar logic to the previous case. Hence $b_{n}=2 a_{n-1}+b_{n-1}$. Therefore, we have $b_{n}=2 a_{n-1}+\frac{1}{2}\left(a_{n}-2 a_{n-1}\right)$, and so $a_{n}=2 a_{n-1}+2 b_{n-1}=2 a_{n-1}+2\left(2 a_{n-2}+\right.$ $\left.\frac{1}{2}\left(a_{n-1}-2 a_{n-2}\right)\right)=3 a_{n-1}+2 a_{n-2}$. Finally, we have $a_{0}=1$ (as the only possibility is to, well, do nothing) and $a_{1}=4$ (as any $2 \times 1$ coloring is admissible), so $a_{2}=14, a_{3}=50, a_{4}=178, a_{5}=634$.