Solution:
Notice that we are solving for $x$ in the domain $-1 < x \leq 2$. Using Cauchy-Schwarz Inequality on the left hand side, one has
$$\sqrt{2 - x} + \sqrt{2 + 2x} = \sqrt{\frac{1}{2} \cdot (4 - 2x) + \sqrt{2 + 2x}} \leq \sqrt{\left(\frac{1}{2} + 1\right)(2 + 2x + 4 - 2x)} = 3.$$
Using the fact that $x^{4} + 1 \geq 2x^{2}$ (because $(x^{2} - 1)^{2} \geq 0$) and $x^{2} + 1 \geq 2x$ (because $(x - 1)^{2} \geq 0$) we can get:
$$2(x^{4} + 1) \geq (x^{2} + 1)^{2} \geq (x^{2} + 1) \frac{(x + 1)^{2}}{2}.$$
If we substitute this into the right-hand side we get
$$\sqrt{\frac{x^{4} + 1}{x^{2} + 1}} + \frac{x + 3}{x + 1} \geq \frac{x + 1}{2} + \frac{2}{x + 1} + 1.$$
Finally applying the AM-GM again for 2 positive numbers $\frac{x + 1}{2}$ and $\frac{2}{x + 1}$ we obtain $\frac{x + 1}{2} + \frac{2}{x + 1} \geq 2$. Thus the right-hand side is $\geq 3$.

Therefore, we can see that equality must occur in our AM-GM inequality and hence, $\frac{x + 1}{2} = \frac{2}{x + 1} \Rightarrow x = 1$.