As
$$1 + \log_2(x^4 + 1) = \log_2(2x^4 + 2),$$
and $\log_2 y$ is monotonically increasing over $(0, +\infty)$, the given inequality is equivalent to
$$x^{12} + 3x^{10} + 5x^8 + 3x^6 + 1 < 2x^4 + 2$$
or
$$x^{12} + 3x^{10} + 5x^8 + 3x^6 - 2x^4 - 1 < 0.$$
It can be rewritten as
$$\begin{aligned} & x^{12} + x^{10} - x^8 \\ & \quad + 2x^{10} + 2x^8 - 2x^6 \\ & \quad + 4x^8 + 4x^6 - 4x^4 \\ & \quad + x^6 + x^4 - x^2 \\ & \quad + x^4 + x^2 - 1 < 0. \end{aligned}$$
That is to say,
$$(x^8 + 2x^6 + 4x^4 + x^2 + 1)(x^4 + x^2 - 1) < 0.$$
Then we have $x^4 + x^2 - 1 < 0$. It follows that $x^2 < \frac{-1+\sqrt{5}}{2}$, i.e.
$$-\sqrt{\frac{-1+\sqrt{5}}{2}} < x < \sqrt{\frac{-1+\sqrt{5}}{2}}.$$
So the solution set is $(-\sqrt{\frac{-1+\sqrt{5}}{2}}, \sqrt{\frac{-1+\sqrt{5}}{2}})$.

As
$$1 + \log_2(x^4 + 1) = \log_2(2x^4 + 2),$$
and $\log_2 y$ is monotonically increasing over $(0, +\infty)$, the given inequality is equivalent to
$$x^{12} + 3x^{10} + 5x^8 + 3x^6 + 1 < 2x^4 + 2$$
or
$$\begin{aligned} \left(\frac{1}{x^2}\right)^3 + 2\left(\frac{1}{x^2}\right) &> x^6 + 3x^4 + 3x^2 + 1 + 2x^2 + 2 \\ & = (x^2 + 1)^3 + 2(x^2 + 1). \end{aligned}$$
Define $g(t) = t^2 + 2t$. Then we have
$$g\left(\frac{1}{x^2}\right) > g(x^2 + 1).$$
Obviously, $g(t)$ is a monotonically increasing function; then we have
$$\frac{1}{x^2} > x^2 + 1.$$
That is to say,
$$x^4 + x^2 - 1 < 0.$$
We obtain $x^2 < \frac{-1 + \sqrt{5}}{2}$. So the solution set is $(-\sqrt{\frac{-1 + \sqrt{5}}{2}}, \sqrt{\frac{-1 + \sqrt{5}}{2}})$.