The expression is equivalent to $\log_{2}\frac{2^{x+1}-3}{4^x+4}=-x$ or $\frac{1}{2^x}=\frac{2^{x+1}-3}{4^x+4}.$ With $t=2^x$ this is equivalent to $t^2-3t-4=0,$ which leads to $t=4$ or $t=-1$. Only the first is acceptable in real numbers, and therefore $x=2$ is the only solution.

I'm going to expand a little bit on tk1's post and make it slightly more understandable. We can subtract the logarithms and use the identity. The motivation behind this is to get the logarithm on one side. Doing anything else would be a tad bit messy and doing this would be more time efficient. We have $x + \log_2(2^{x+1} -3) - \log_2(4^x +4) = 0$. Rewriting this through the logarithm identity we have $x + \log_2(\frac{2^{x+1} -3}{4^x + 4}) = 0$. Let's subtract $x$ and raise everything to the power of $2$ to make it less messy. We now have $\frac{2^{x+1} -3}{4^x +4} = \frac{1}{2^x}$. Working with this stuff without variables would be messy so we can set, arbitrarily, $t = 2^x$. $\frac{2 * t - 3}{(2^2)^x +4} = \frac{1}{t}$. The $(2^2)^x$ can be rewritten as $2^x * 2^x = t^2$. Let's cross multiply to get $2t^2 - 3t = t^2 + 4$. Which turns into the quadratic $t^2 - 3t -4 = 0$. We factor this to get $(t-4)(t+1) = 0$. This means that $t = 4, -1$. But since we are looking in the real numbers, $t = -1$ is turned invalid. So we have $2^x = 4$, implying that $x = 2$ $ \fbox{x=2}$.