6. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n}^{2}=a_{n+1} a_{n}-1\left(n \in \mathbf{Z}_{+}\right) \text {, and } a_{1}=\sqrt{2} \text {. }
$$

Then the natural number closest to $\sqrt{a_{2014}}$ is $\qquad$
6.8.

From the given, we have
$$
\begin{array}{l}
a_{n+1}=a_{n}+\frac{1}{a_{n}} \Rightarrow a_{n+1}^{2}-a_{n}^{2}=2+\frac{1}{a_{n}^{2}} \\
\Rightarrow a_{n+1}^{2}=a_{1}^{2}+2 n+\sum_{i=1}^{n} \frac{1}{a_{i}^{2}} .
\end{array}
$$

Thus, $a_{2014}^{2}=2+2 \times 2013+\sum_{i=1}^{2013} \frac{1}{a_{i}^{2}}$
$$
>2+2 \times 2013=4028>3969=63^{2} \text {. }
$$

Also, $a_{1}=\sqrt{2}, a_{n+1}=a_{n}+\frac{1}{a_{n}}>2\left(n \in \mathbf{Z}_{+}\right)$, so
$$
\begin{array}{l}
a_{2014}^{2}=2+2 \times 2013+\frac{1}{2}+\sum_{i=2}^{2013} \frac{1}{a_{i}^{2}} \\
<4028+\frac{1}{2}+\frac{1}{4} \times 2012=4531.5 \\
<4624=68^{2} .
\end{array}
$$

Therefore, $9 \approx \sqrt{63}<\sqrt{a_{2014}}<\sqrt{68} \approx 8.3$.
Thus, the natural number closest to $\sqrt{a_{2014}}$ is 8.