Find value range of $2^x+2^y$ Assume $x,y \in \Bbb{R}$ satisfy $$4^x+4^y = 2^{x+1} + 2^{y+1}$$, Find the value range of $$2^x+2^y$$

I know $x=y=1$ is a solution of $4^x+4^y = 2^{x+1} + 2^{y+1}$ , but I can't go further more. I can only find one solution pair of $4^x+4^y = 2^{x+1} + 2^{y+1}$. It seems very far from solve this question... 

So, then the another solution:
let $f(x) = 2^x$. Then we can go to that form of equation:
$$(f(x)-1)^2 + (f(y)-1)^2 = 2$$
We got a circle. If we consider points of intersection of this circle with a lines $$f(x)+f(y) = const$$ we can get an answer.
We should also be polite with restrictions $f(x),f(y) > 0$.
If we draw a picture, we will conclude that range is (2, 4].
