(1) If a positive integer $n$ can be expressed in the form $a^{b} (a, b \in \mathbf{N}, a, b \geqslant 2)$, then $n$ is called a "good number." Find all good numbers adjacent to positive integer powers of 2.
(2) Find all non-negative integer solutions $(x, y, z)$ of the indeterminate equation $\left|2^{x}-3^{y} \times 5^{x}\right|=1$.
(1) Let the required good number be $n, n=a^{b}(a, b \in \mathbf{N}_{+}, a \geqslant 2, b \geqslant 2)$.
Thus, there exists a positive integer $t(t>1)$, such that
$2^{t}=a^{b} \pm 1$.
Clearly, $a$ is an odd number.
If $b$ is odd, then
$$
2^{t}=(a \pm 1)\left(a^{b-1} \mp a^{b-2}+\cdots \mp a+1\right).
$$

And $a^{b-1} \mp a^{b-2}+\cdots \mp a+1$ is the result of adding and subtracting an odd number of odd numbers, which is still odd, and can only be 1. Substituting into equation (1) gives $b=1$, which contradicts $b \geqslant 2$.
If $b$ is even, then $a^{b} \equiv 1(\bmod 4)$.
If $2^{t}=a^{b}+1$, then $2^{t}=a^{b}+1 \equiv 2(\bmod 4)$.
Therefore, $t=1$. Contradiction.
If $2^{t}=a^{b}-1=\left(a^{\frac{b}{2}}+1\right)\left(a^{\frac{b}{2}}-1\right)$, but $\left(a^{\frac{6}{2}}+1, a^{\frac{6}{2}}-1\right)=2$, hence
$a^{\frac{b}{2}}-1=2 \Rightarrow a^{b}=9$.
In summary, the only good number sought is $n=9$.
(2) Clearly, $x \geqslant 1$.
When $z=0$,
If $y \leqslant 1$, it is easy to find three solutions of the equation
$$
(1,0,0),(1,1,0),(2,1,0);
$$

If $y \geqslant 2$, by the conclusion of (1), it is easy to know that the equation has only one solution $(3,2,0)$ at this time.
When $z \geqslant 1$, clearly, $x \geqslant 2$. It is easy to know
When and only when $x \equiv 2(\bmod 4)$,
$2^{x} \equiv-1(\bmod 5)$;
When and only when $x \equiv 0(\bmod 4)$,
$2^{x} \equiv 1(\bmod 5)$.
If $2^{x}-3^{y} \times 5^{x}=1$,

then $2^{x} \equiv 1(\bmod 5)$, at this time, $x \equiv 0(\bmod 4)$.
Let $x=4 m\left(m \in \mathbf{N}_{+}\right)$.
Taking both sides of equation (2) modulo 4, we get
$(-1)^{\gamma+1} \equiv 1(\bmod 4)$.
Thus, $y$ is odd. Let $y=2 l+1(l \in \mathbf{N})$.
Then equation (2) becomes $2^{4 m}-3^{2 l+1} \times 5^{x}=1$, i.e.,
$\left(2^{2 m}-1\right)\left(2^{2 m}+1\right)=3^{2 l+1} \times 5^{3}$.
By $\left(2^{2 m}-1,2^{2 m}+1\right)=1,31\left(2^{2 m}-1\right)$, we have
$\left\{\begin{array}{l}2^{2 m}-1=3^{2 l+1}, \\ 2^{2 m}+1=5^{x} .\end{array}\right.$
Combining the conclusion of (1), the only $(m, l)$ that satisfies equation (3) is $(1,0)$, substituting into equation (4) gives $z=1$.
At this time, one solution of the original equation is $(4,1,1)$.
If $3^{y} \times 5^{x}-2^{x}=1$,

then $2^{x} \equiv-1(\bmod 5)$, at this time, $x \equiv 2(\bmod 4)$.
Let $x=4 k+2(k \in \mathbf{N})$. Then
$3^{y} \times 5^{x}=2^{4 k+2}+1$.
When $k=0$, $y=0, z=1$, one solution of the original equation is $(2,0,1)$.
When $k \geqslant 1$, taking both sides of equation (6) modulo 4, we get
$(-1)^{y} \equiv 1(\bmod 4)$.
Thus, $y$ is even. Let $y=2 r(r \in \mathbf{N})$.
At this time, taking both sides of equation (6) modulo 8, we get
$5^{x} \equiv 1(\bmod 8)$.
Thus, $z$ is even. Let $z=2 s(s \in \mathbf{N})$.
Thus, equation (6) becomes
$\left(3^{\prime} \times 5^{\prime}\right)^{2}-1=2^{4++2}$.
Combining the conclusion of (1), we know $3^{r} \times 5^{\prime}=3$.
Thus, $2^{4 k+2}=8$, contradiction.
Therefore, $(x, y, z)=(1,0,0),(1,1,0),(2,1,0)$,
$$
(3,2,0),(4,1,1),(2,0,1).
$$