10. (20 points) Given the sequence $\left\{a_{n}\right\}:$
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, \\
\frac{a_{n+2}}{a_{n}}=\frac{a_{n+1}^{2}+1}{a_{n}^{2}+1}(n=1,2, \cdots) .
\end{array}
$$

Find $\left[a_{2021}\right]$.
10. From the recurrence relation, we have
$$
\prod_{i=1}^{n-2} \frac{a_{i+2}}{a_{i}}=\prod_{i=1}^{n-2} \frac{a_{i+1}^{2}+1}{a_{i}^{2}+1} .
$$

Simplifying, we get $\frac{a_{n} a_{n-1}}{a_{1} a_{2}}=\frac{a_{n-1}^{2}+1}{a_{1}^{2}+1}$.
Given $a_{1}=1, a_{2}=2$, we know
$$
a_{n}=a_{n-1}+\frac{1}{a_{n-1}} \text {. }
$$

On one hand,
$$
\begin{aligned}
a_{n}^{2} & =a_{n-1}^{2}+2+\frac{1}{a_{n-1}^{2}} \\
& >a_{n-1}^{2}+2>\cdots>a_{1}^{2}+2(n-1),
\end{aligned}
$$

it is easy to see that $a_{2021}^{2}>4041$, hence $a_{2021}>63$.
On the other hand, from
$$
\begin{array}{l}
a_{n}^{2}=a_{n-1}^{2}+2+\frac{1}{a_{n-1}^{2}}=\cdots \\
=a_{1}^{2}+2(n-1)+\sum_{i=1}^{n-1} \frac{1}{a_{i}^{2}}, \\
\text { then } a_{2021}^{2}=4041+\sum_{i=1}^{2020} \frac{1}{a_{i}^{2}} .
\end{array}
$$

From equation (1), we know
$$
\begin{array}{l}
a_{n}>a_{n-1}, \\
\sum_{i=1}^{2020} \frac{1}{a_{i}^{2}}=\sum_{i=1}^{10} \frac{1}{a_{i}^{2}}+\sum_{i=11}^{100} \frac{1}{a_{i}^{2}}+\sum_{i=101}^{2020} \frac{1}{a_{i}^{2}} \\
<\frac{10}{a_{1}^{2}}+\frac{90}{a_{10}^{2}}+\frac{1920}{a_{100}^{2}}<10+\frac{90}{19}+\frac{1920}{199} \\
<25 \\
\Rightarrow a_{2021}^{2}<4041+25<4096=64^{2} \\
\Rightarrow a_{2021}<64 .
\end{array}
$$

In summary, $\left[a_{2021}\right]=63$.