Romanian Mathematical Olympiad

The equation can be written $\log_2(1 + \sqrt{x}) = (2^x - 1)^2$.
The function $f : [0, \infty) \to [0, \infty)$, defined by $f(x) = \log_2(1 + \sqrt{x})$, is one to one and onto and its inverse $f^{-1} : [0, \infty) \to [0, \infty)$ is given by $f^{-1}(x) = (2^x - 1)^2$.
As $f$ is increasing strictly, the equation can be written $f(x) = f^{-1}(x) = x$, or equivalently $2^x = 1 + \sqrt{x}$. One can easily check that $x_1 = 0$ and $x_2 = 1$ are solutions.
As $g : [0, \infty) \to [1, \infty)$, $g(x) = 2^x$ is convex and the function $h : [0, \infty) \to [1, \infty)$, $h(x) = 1 + \sqrt{x}$ is concave, these are the only solutions.