Let $ a = \sqrt {2 + \sqrt {3}}$. Note that $ 1 < a < 2$. We have: $ a^{2x} + 1 = (2a)^x \Longleftrightarrow a^x + a^{ - x} = 2^x$. Since $ 2^x = a^x + a^{ - x} \ge 2$, we have $ x \ge 1$. Let $ f(x) = a^x + a^{ - x}$ and $ g(x) = 2^x$. Note that $ f'(x) = \ln(a)(a^{x} - a^{ - x})$ and $ g'(x) = \ln(2)2^x$. Since $ 0 < \ln(a) < \ln(2)$, and $ 0 < a^{x} - a^{ - x} < a^x < 2^x$ for all $ x \ge 0$, we have $ f'(x) < g'(x)$ for all $ x \ge 0$. Since $ 2 = f(0) > g(0) = 1$, and $ f'(x) < g'(x)$ for all $ x \ge 0$, there must be exactly one solution to this equation. By inspection $ x = 2$ yields: $ f(2) = (2 + \sqrt {3}) + (2 - \sqrt {3}) = 4$, and $ g(2) = 2^2 = 4$. Thus, the only solution is $ x = 2$.