First, find constraints for base and argument:

1. Base $x^2 > 0$ and $x^2 \neq 1 \implies x \neq 0$ and $x \neq \pm 1$.
2. Argument $x + 2 > 0 \implies x > -2$.

Now, analyze the inequality in two cases based on the value of the base $x^2$:

Case 1: $0 < x^2 x^2 \implies x^2 - x - 2 < 0 \implies (x-2)(x+1) < 0 \implies -1 < x 1 \implies x \in (-\infty, -1) \cup (1, \infty)$. Since the base is greater than 1, the inequality sign remains the same: $x+2 0 \implies x 2$. Intersection with $x > -2$ and base constraints for Case 2 yields: $x \in (-2, -1) \cup (2, \infty)$.

Combining both cases, we get: $x \in (-2, -1) \cup (-1, 0) \cup (0, 1) \cup (2, \infty)$.