Four circles of radius 1 are each tangent to two sides of a square and externally tangent to a circle of radius 2, as shown.  What is the area of the square?

[asy]
unitsize(1cm);draw(Circle((0,0),2));
for(int i=0; i<4; ++i) {
	draw(Circle(scale(3)*dir(45+90*i),1));
	draw((3+sqrt(2))*dir(45+90*i)--(3+sqrt(2))*dir(-45+90*i));  
}
draw((0,0)--(2,0));
label("2",(1,0),N);
draw(scale(3)*dir(45+90*1)--shift((1,0))*scale(3)*dir(45+90*1));
label("1",point(scale(3)*dir(45+90*1)--shift((1,0))*scale(3)*dir(45+90*1),.5),S);
[/asy]
Let $s$ be the length of a side of the square. Consider an isosceles right triangle with vertices at the centers of the circle of radius 2 and two of the circles of radius 1. This triangle has legs of length 3, so its hypotenuse has length $3\sqrt{2}$.

[asy]
unitsize(1cm);
draw(Circle((0,0),2));
for(int i=0; i<4; ++i) {
	draw(Circle(scale(3)*dir(45+90*i),1));
	draw((3+sqrt(2))*dir(45+90*i)--(3+sqrt(2))*dir(-45+90*i));  
}
pair A = scale(3)*dir(45), B = scale(3)*dir(45+90);
draw(A--B--origin--cycle);
label("$1$", A, SE);
label("$1$", B, SW);
label("$2$", point(origin--A,.3), SE);
label("$2$", point(origin--B,.3), SW);
draw(rightanglemark(A,origin,B,5));
[/asy]

The length of a side of the square is 2 more than the length of this hypotenuse, so  $s=2 + 3\sqrt{2}$. Hence the area of the square is \[
s^{2}=(2+3\sqrt{2})^{2}=\boxed{22+12\sqrt{2}}.
\]