Three. (50 points) In a $\left(2^{n}-1\right) \times\left(2^{n}-1\right)(n$ $\geqslant 2)$ grid, each cell is filled with 1 or -1. If the number in any cell is equal to the product of the numbers in the cells that share an edge with it, then this filling method is called "successful". Find the total number of "successful" fillings.

In a $\left(2^{n}-1\right) \times\left(2^{n}-1\right)(n$ $\geqslant 2)$ grid, each cell is filled with 1 or -1. If the number in any cell is equal to the product of the numbers in the cells that share an edge with it, then this filling method is called "successful". Find the total number of "successful" fillings.
Three, assuming there exists some successful filling method that contains -1. First, prove: if this successful filling method is symmetric about the middle column (row), then the middle column (row) is entirely 1.
Let $a_{0}=1, a_{2}^{n}=1$.
If $a_{1}=1$, by $a_{1}=a_{0} \times a_{2} \times 1$, we get $a_{2}=1$.

Similarly, $a_{3}=1, \cdots$, $a_{2^{n}-1}=1$.
If $a_{2}^{n}-1=1$, similarly, we get $a_{1}=a_{2}=\cdots=a_{2}-2=1$.
Therefore, if $a_{1}=-1$, then $a_{2}^{n}-1=-1$. This way,
$$
\begin{array}{l}
a_{1}=-1, a_{2}=-1, a_{3}=1, a_{4}=-1, a_{5}=-1, a_{6} \\
=1, \cdots, a_{2}^{n}-2=-1, a_{2}^{n}-1=-1, a_{2}^{n}=1 \text {. Thus } 312^{2},
\end{array}
$$

a contradiction.
Therefore, the middle column (row) is entirely 1.
Secondly, if this successful filling method is not symmetric about the middle column, first rotate this filling method 180° along the middle column, to get another filling method, then multiply the numbers in the same positions of the two filling methods, to get a successful filling method symmetric about the middle column, which contains -1. This successful filling method can be further transformed into a successful filling method that is symmetric about both the middle row and the middle column, which contains -1.

For such a $\left(2^{n}-1\right) \times\left(2^{n}-1\right)$ successful filling method, removing all 1s in the middle row and the middle column, we can get 4 $\left(2^{n-1}-1\right) \times\left(2^{n-1}-1\right)$ successful filling methods, at least one of which contains -1. For this successful filling method, repeat the above operation, and eventually, we can get a $3 \times 3$ successful filling method, which contains -1, but the middle row and the middle column are all 1, which contradicts the definition of a successful filling method.

Therefore, for any successful filling method, it must not contain -1. Hence, there is only 1 successful filling method, which is to fill each cell with 1.
(Anqing No.1 High School, Anhui Province, 246004)