Hmm.
[tex]\frac{d}{dx} 2^x = \ln 2 2^x[/tex]
which is monotone increasing and
[tex]\frac{d}{dx} x = 1[/tex]
which is constant, so if
[tex]2^x > x[/tex]
where
[tex]\ln 2 \times 2^x = 1 \rightarrow x= -\log_2 ({\ln 2}) \approx 0.5[/tex]
then the only solutions are imaginary.
So I don't think there are any real solutions.

2^(2^x) does not equal x, it equals 2^x; you have to do the same thing to both sides, right??? I guess I don't understand your question then. It definitely does not have real solutions, because of the reason above.