Example 10 Given a natural number $a \geqslant 2$, the sets are
$$
\begin{array}{l}
A=\left\{y \mid y=a^{x}, x \in \mathbf{N}\right\}, \\
B=\{y \mid y=(a+1) x+b, x \in \mathbf{N}\} .
\end{array}
$$

Does there exist a $b$ in the interval $[1, a]$ such that $C=A \cap B$ $\neq \varnothing$? If it exists, find all possible values of $b$ and the corresponding set $C$; if not, explain the reason.
$$
\begin{array}{l}
\text { Solution: } \because a \geqslant 2, a \in \mathbf{N}, x \leqslant \mathbf{N}, \\
\therefore a \in \mathbf{N}, \text { only if } (a+1) x \in \mathbf{N} . \\
\text { Since } A \cap B \cdot 7^{\prime}-6, \\
\therefore z \in \mathbf{N} .
\end{array}
$$

According to the problem, we only need to find $b \in \mathbf{N}$ such that
$$
(a+1) x_{2}+b=a^{x_{1}},
$$

i.e., $x_{2}=\frac{a^{x_{1}}-1}{a+1}, x_{1} 、 x_{2} \in \mathbf{N}$.
When $x_{1}$ is a positive even number,
$$
\begin{array}{l} 
x_{2}=\frac{a^{x_{1}}-b}{a+1}=\frac{-\left(1-a^{x_{1}}\right)-b+1}{1+a} . \\
=-\frac{1-a^{x_{1}}}{1+a}-\frac{b-1}{1+a} . \\
\because-\frac{1-a^{x_{1}}}{1+a} \in \mathbf{N}, 1 \leqslant b \leqslant a, \\
\therefore 0 \leqslant \frac{b-1}{1+a}<1 . \\
\because x_{2} \in \mathbf{N}, \therefore b=1 .
\end{array}
$$

When $x_{1}$ is a positive odd number,
$$
\begin{array}{l} 
x_{2}=\frac{a^{x_{1}}-b}{a+1}=\frac{1+a^{x_{1}}-b-1}{1+a} \\
=\frac{a^{x_{1}}+1}{a+1}-\frac{b+1}{a+1} . \\
\because \frac{a^{x_{1}}+1}{a+1} \text { is a natural number greater than } 1, 0 \leqslant \frac{b+1}{a+1} \\
, x_{2} \in \mathbf{N}, \\
\therefore b=a .
\end{array}
$$
$\because \frac{a^{x_{1}}+1}{a+1}$ is a natural number greater than 1, $0 \leqslant \frac{b+1}{a+1}$ $\leqslant 1, x_{2} \in \mathbf{N}$,

In summary, there exists $b=1$ or $b=a$ in $[1, a]$ such that $A \cap B \neq \varnothing$.
When $b=1$, from the set $A$ we get
$$
A \cap B=\left\{y \mid y=a^{2 x}, x \in \mathbf{N}\right\} ;
$$

When $b=a$, we get
$$
A \cap B=\left\{y \mid y=a^{2 x+1}, x \in \mathbf{N}\right\} .
$$