14. Given a sequence $\left\{a_{n}\right\}$ with all terms being positive, the sum of the first $n$ terms is $S_{n}$, and $a_{n}+1=2 \sqrt{S_{n}}\left(n \in \mathbf{Z}_{+}\right)$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=\frac{a_{n+2}}{a_{n} a_{n+1} 2^{n}}$, the sum of the first $n$ terms of the sequence $b_{n}$ is $T_{n}$, prove: $\frac{5}{6} \leqslant T_{n}<1$.
14. (1) From $a_{n}+1=2 \sqrt{S_{n}}$, we know that when $n=1$,
$$
a_{1}+1=2 \sqrt{S_{1}}=2 \sqrt{a_{1}} \Rightarrow S_{1}=1 \text {; }
$$

When $n \geqslant 2$,
$$
\begin{array}{l}
a_{n}=S_{n}-S_{n-1}=2 \sqrt{S_{n}}-1 \\
\Rightarrow\left(\sqrt{S_{n}}-1\right)^{2}=S_{n-1} \\
\Rightarrow \sqrt{S_{n}}=\sqrt{S_{n-1}}+1\left(a_{n}>0\right) .
\end{array}
$$

Thus, $\left\{\sqrt{S_{n}}\right\}$ is an arithmetic sequence with the first term 1 and common difference 1.
Therefore, $\sqrt{S_{n}}=n, a_{n}=2 \sqrt{S_{n}}-1=2 n-1$.
(2) From (1), we know $a_{n}=2 n-1$.
$$
\begin{array}{l}
\text { Then } b_{n}=\frac{2 n+3}{(2 n-1)(2 n+1) 2^{n}} \\
=\left(\frac{2}{2 n-1}-\frac{1}{2 n+1}\right) \frac{1}{2^{n}} \\
=\frac{1}{2^{n-1}(2 n-1)}-\frac{1}{2^{n}(2 n+1)} .
\end{array}
$$

Thus, $T_{n}=b_{1}+b_{2}+\cdots+b_{n}$
$$
\begin{aligned}
= & 1-\frac{1}{6}+\frac{1}{6}-\frac{1}{20}+\cdots+\frac{1}{2^{n-1}(2 n-1)}- \\
& \frac{1}{2^{n}(2 n+1)} \\
= & 1-\frac{1}{2^{n}(2 n+1)} .
\end{aligned}
$$

Since $\frac{1}{2^{n}(2 n+1)}>0$, it is clear that $T_{n}<1$.
Also, $T_{n}=1-\frac{1}{2^{n}(2 n+1)}$ is monotonically increasing, thus,
$$
T_{n} \geqslant T_{1}=\frac{5}{6} \text {. }
$$

In summary, $\frac{5}{6} \leqslant T_{n}<1$.