14. Let $a \in \mathbf{N}_{+}, a \geqslant 2$, and the sets $A=\left\{y \mid y=a^{x}\right.$, $\left.x \in \mathbf{N}_{+}\right\}, B=\left\{y \mid y=(a+1) x+b, x \in \mathbf{N}_{+}\right\}$. Does there exist $b$ in the closed interval $[1, a]$ such that $A \cap B \neq \varnothing$? If it exists, find all possible values of $b$ and the corresponding $A \cap B$; if not, explain the reason.
14. Let $b \in [1, a]$ such that $A \cap B \neq \varnothing$, i.e., there exists $y_{0} \in A$ and $y_{0} \in B$, such that $y_{0} = a^{m}$ ($m \in \mathbf{N}_{+}$) and $y_{0} = (a+1)n + b$ ($n \in \mathbf{N}_{+}$). Then there should exist $m, n \in \mathbf{N}_{+}$ such that $a^{m} = (a+1)n + b$ ($1 \leqslant b \leqslant a$),
i.e.,
$$
\begin{aligned}
n = & \frac{a^{m} - b}{a + 1} \left(m, n \in \mathbf{N}_{+}, 1 \leqslant b \leqslant a\right). \\
\because & a^{m} - b = [(a+1) - 1]^{m} - b \\
& = (a+1)^{m} - C_{m}^{1}(a+1)^{m-1} + \cdots \\
& + C_{m}^{m-1}(-1)^{m-1}(a+1) + (-1)^{m} - b,
\end{aligned}
$$
$\therefore n \in \mathbf{N}_{+}$ when $(-1)^{m} - b$ is divisible by $a+1$.
(1) When $m$ is a positive even number, $1 - b$ should be divisible by $a+1$. Since $a \geqslant 2, 1 \leqslant b \leqslant a$, it is only satisfied when $b = 1$.
(2) When $m$ is a positive odd number, $-1 - b$ should be divisible by $a+1$. Since $2 \leqslant b+1 \leqslant a+1$, it is only satisfied when $b = a$.

In summary, the $b$ that satisfies the condition exists, and its values are $b = 1$ or $b = a$.
When $b = 1$, $A \cap B = \left\{y \mid y = a^{2k}, k \in \mathbf{N}_{+}\right\}$;
When $b = a$, $A \cap B = \left\{y \mid y = a^{2k+1}, k \in \mathbf{N}_{+}\right\}$.