(1) When $b < 0$ and $m$ is even, in order that $a b^{m-1} < -2$, we should first have $a b^m + b > -b > 0$, and therefore $a(a b^m + b)^m + b > a b^m + b > 0$, i.e. $x_3 > x_2 > 0$. Using the fact that $a x^m + b$ is monotonically increasing on $(0, +\infty)$, it can be established that each succeeding term of the sequence $\{x_n\}$ is greater than its preceding term, and is greater than $-b$ starting from the second term.

Considering any three consecutive terms of the sequence $x_n, x_{n+1}, x_{n+2}, \dots$, we have
$$\begin{aligned} x_{n+2} - x_{n+1} &= a(x_{n+1}^m - x_n^m) \\ &= a(x_{n+1} - x_n)(x_{n+1}^{m-1} + x_{n+1}^{m-2} x_n + \dots + x_n^{m-1}) \\ &> a m x_n^{m-1}(x_{n+1} - x_n) \\ &> a m (-b)^{m-1}(x_{n+1} - x_n) \\ &> 2 m (x_{n+1} - x_n) \\ &> x_{n+1} - x_n. \end{aligned}$$
It is obvious that the difference of any two consecutive terms of the sequence $\{x_n\}$ is increasing, and hence it is not bounded.

When $a b^{m-1} \ge -2$, mathematical induction is used to prove that each term of the sequence $\{x_n\}$ falls on the interval $[b, -b]$.

The first term $b$ falls on the interval $[b, -b]$. Suppose that the term $x_n$ satisfies the condition $b \le x_n \le -b$ for a particular $n$. Then $0 \le x_n^m \le b^m$, and hence
$$b = a \times 0^m + b \le x_{n+1} \le a b^m + b \le -b.$$
Thus, the sequence $\{x_n\}$ is bounded if and only if $a b^{m-1} \ge -2$.

(2) When $b > 0$, each term of the sequence $\{x_n\}$ is positive. So, we first prove that $\{x_n\}$ is bounded if and only if the equation $a x^m + b = x$ has positive real roots.

Suppose that $a x^m + b = x$ has no positive real roots. In such a case, the minimum value of the function $p(x) = a x^m + b - x$ on the interval $(0, +\infty)$ is greater than zero. Let $t$ be the minimum value. It follows that for any two consecutive terms of the sequence $x_n$ and $x_{n+1}$, we have $x_{n+1} - x_n = a x_n^m - x_n + b$. Thus, each succeeding term of the sequence $\{x_n\}$ is greater than the preceding term by at least $t$. Hence, it is not bounded.

If the equation $a x^m + b = x$ has positive real roots, let $x_0$ be one of the positive real roots. Then, by using mathematical induction, we prove that each term of the sequence $\{x_n\}$ is less than $x_0$. Firstly, the first term $b$ is less than $x_0$. Suppose that $x_n < x_0$ for a particular $n$. By virtue of the fact that $a x^m + b$ is increasing on the interval $[0, +\infty)$, it can be established that
$$x_{n+1} = a x_n^m + b < a x_0^m + b = x_0.$$
Therefore, the sequence is bounded.

Further, the equation $a x^m + b = x$ has positive roots if and only if the minimum value of $a x^{m-1} + \frac{b}{x}$ on the interval $(0, +\infty)$ is not greater than $1$, whereas the minimum value of $a x^{m-1} + \frac{b}{x}$ can be determined by mean inequality, i.e.
$$a x^{m-1} + \frac{b}{x} = a x^{m-1} + \frac{b}{(m-1)x} + \dots + \frac{b}{(m-1)x} \geq m \sqrt[m]{\frac{a b^{m-1}}{(m-1)^{m-1}}}$$
As such, the sequence $\{x_n\}$ is bounded if and only if
$$m \sqrt[m]{\frac{a b^{m-1}}{(m-1)^{m-1}}} \le 1, \text{ i.e. } a b^{m-1} \le \frac{(m-1)^{m-1}}{m^m}.$$

When $b < 0$, and $m$ is odd, let $y_n = -x_n$. Then $y_1 = -b > 0$, $y_{n+1} = a y_n^m + (-b)$, showing that the sequence $\{x_n\}$ is bounded if and only if the sequence $\{y_n\}$ is bounded. Thus, by using the above reasoning, it can be proven that (2) holds.