Answer
$\{\log_{4}(-2+\sqrt{7})\}$ or $\{-0.315\}$
Work Step by Step
...$ \quad 16^{x}=(4^{2})^{x}=(4^{x})^{2}$
...$ \quad 4^{x+1}=4^{x}\cdot 4$
... Substitute $t=4^{x}$
... the equation becomes
$t^{2}+4t-3=0 \quad $...$ \quad $ quadratic formula...
$t=\displaystyle \frac{-4\pm\sqrt{4^{2}-4(1)(-3)}}{2(1)}=\frac{-4\pm\sqrt{28}}{2}=\frac{-4\pm 2\sqrt{7}}{a}=-2\pm\sqrt{7}$
$t=4^{x}$ is positive so we discard the negative t,
$t=-2+\sqrt{7}$
Now, bringing x back,
$4^{x}=-2+\sqrt{7} \quad $...Apply$: \quad \log_{4}(...)$
$x=\log_{4}(-2+\sqrt{7})\approx-0.315$
The solution set is $\{\log_{4}(-2+\sqrt{7})\}$ or $\{-0.315\}$
