Number of binary square matrices whose every contiguous 2 by 2 submatrix has exactly 2 ones (zeroes) 
What is the number of binary $n\times n$ matrices whose every contiguous  $2\times 2$ submatrix has exactly two ones (zeroes)?

That is, if $a_{ij}\in\{0,1\}$ for $i,j\in\{1,2,\dots,n\}$ are matrix entries, then $$a_{i,j} + a_{i,j+1} + a_{i+1,j} + a_{i+1,j+1}= 2$$
for $i \in \{1,\dots,n-1\}$ and $j \in \{1,\dots,n-1\}$, where $n\ge 2$.
After writing out some cases for small $n$, it appears to me that the answer might be $2^{n+1}-2$.
Is there an elegant way to prove this?

For example, for $n=2$ we have $6$ matrices
$$
\begin{bmatrix}
1 &0 \\
1 &0 \\
\end{bmatrix}, 
\begin{bmatrix}
1 &1 \\
0 &0 \\
\end{bmatrix}, 
\begin{bmatrix}
1 &0 \\
0 &1 \\
\end{bmatrix}, 
\begin{bmatrix}
0 &1 \\
1 &0 \\
\end{bmatrix}, 
\begin{bmatrix}
0 &0 \\
1 &1 \\
\end{bmatrix}, 
\begin{bmatrix}
0 &1 \\
0 &1 \\
\end{bmatrix}.
$$
For example, for $n=3$ we have $14$ matrices:

*

*$4$ rotations of
$\begin{bmatrix}
1 &0 &1 \\
1 &0 &1 \\
0 &1 &0 \\
\end{bmatrix}$, $2$ rotations of
$\begin{bmatrix}
1 &0 &1 \\
1 &0 &1 \\
1 &0 &1 \\
\end{bmatrix}$, $1$ matrix
$\begin{bmatrix}
1 &0 &1 \\
0 &1 &0 \\
1 &0 &1 \\
\end{bmatrix}$.

*One additional $(J-M)$ matrix for every of $7$ matrices $M$ from previous case.

$J$ is a matrix filled with $1$'s at every entry.

The first row can be any of the $2^n$ members of $\{0,1\}^n$.
Given the $k$'th row $(x_1, \ldots, x_n)$, the $k+1$'th row can always be
$(1-x_1, \ldots, 1-x_n)$.  The only other possibility is that the $k+1$'th row is
$(x_1,\ldots, x_n)$, but only if the $k$'th row is alternating $(0,1,0,1,\ldots)$ or $(1,0,1,0,\ldots)$. Now use induction on the number of rows...
