Three. (50 points) Given that $n$ is a positive integer not exceeding 2021, and satisfies
$$
\left([\sqrt{n}]^{2}+1\right) \mid\left(n^{2}+1\right) \text {. }
$$

Find the number of such $n$.
$$
\begin{array}{l}
\text { (3) When } n=m^{2}\left(m \in \mathbf{Z}_{+}\right), \text { equation (1) becomes } \\
(n+1) \mid\left(n^{2}+1\right) \text {. } \\
\text { By }\left(n+1, n^{2}+1\right)=(n+1,1-n) \\
=(n+1,2) \leqslant 2 \text {, } \\
\text { and } n+1 \geqslant 2 \text {, thus, } n+1=2 \text {. } \\
\end{array}
$$

Therefore, $n=1$, which satisfies equation (1).
(2) Suppose $m^{2}m^{2}=2 \times 288 \times 289>2021$ (discard).
Thus, when $l \geqslant 2$, the system of equations (2) has no solutions that meet the conditions.
From the system of equations (3), we know $u^{2}-2 v^{2}=2$.
Then $2 \mid u$, let $u=2 s$.
Thus, $v^{2}-2 s^{2}=-1$, which is also a basic form of the Pell equation, and its smallest positive solution is $(v, s)=(1,1)$, and its positive general solution is
$$
\begin{array}{l}
v+s \sqrt{2}=(1+\sqrt{2})^{2 l-1}\left(l \in \mathbf{Z}_{+}\right) . \\
\text { If } l=1 \text {, then } v=s=1 \text {, so } \\
u=2, k=1 \Rightarrow m^{2}=4 \\
\Rightarrow a=4 \Rightarrow n=8 .
\end{array}
$$

Upon verification, $n=8$ satisfies equation (1).
If $l=2$, then
$$
\begin{array}{l}
v+s \sqrt{2}=(1+\sqrt{2})^{3}=7+5 \sqrt{2} \\
\Rightarrow v=7, s=5 \\
\Rightarrow u=10, k=v^{2}=49 .
\end{array}
$$

Then $n>m^{2}=2 \times 49 \times 50>2021$ (discard).
Thus, when $l \geqslant 2$, the system of equations (3) also has no solutions that meet the conditions.

When $k=3$, $(a-1)^{2}+1=3\left(m^{2}+1\right)$, and checking the equation modulo 3 shows it does not hold.
In summary, the number of $n$ that satisfy the conditions is
$1+44+1+1=47$.
$$