Three, (50 points) Find the smallest two positive integers $m$, such that $47\left(m^{2}+46 m+713\right)$ is a perfect square.

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Three, note that,
$$
\begin{array}{l}
47\left(m^{2}+46 m+713\right) \\
=47[47(m+15)+m(m-1)+8] .
\end{array}
$$

Thus, the necessary condition for $47\left(m^{2}+46 m+713\right)$ to be a perfect square is
$$
47 \mid[m(m-1)+8] \text {. }
$$

Let $m=47 k+i(k=0,1, \cdots ; i=0,1, \cdots, 46)$.
Then $m(m-1)+8$
$$
\begin{array}{l}
=(47 k+i)(47 k+i-1)+8 \\
=47 k(47 k+2 i-1)+i(i-1)+8 .
\end{array}
$$

Hence, $47 \mid[m(m-1)+8]$
$$
\Leftrightarrow 47 \mid[i(i-1)+8] \Leftrightarrow i=22,26 \text {. }
$$

Therefore, $47\left(m^{2}+46 m+713\right)$ can only be a perfect square when
$$
m=22(\bmod 47) \text { or } m=26(\bmod 47)
$$

When $m=47 k+22(k=0,1, \cdots)$,
$$
47\left(m^{2}+46 m+713\right)=47^{2}\left(47 k^{2}+90 k+47\right) \text {. }
$$

When $m=47 k+26(k=0,1, \cdots)$,
$$
47\left(m^{2}+46 m+713\right)=47^{2}\left(47 k^{2}+98 k+55\right) \text {. }
$$

When $k$ is even,
$$
47 k^{2}+90 k+47,47 k^{2}+98 k+55
$$

are not perfect squares.
If $k=2 n-1\left(n \in \mathbf{Z}_{+}\right)$, then
$$
\begin{array}{l}
47 k^{2}+90 k+47=4\left(47 n^{2}-2 n+1\right), \\
47 k^{2}+98 k+55=4\left(47 n^{2}+2 n+1\right) .
\end{array}
$$

Thus, the necessary condition for $47\left(m^{2}+46 m+713\right)$ to be a perfect square is that $47 n^{2} \pm 2 n+1\left(n \in \mathbf{Z}_{+}\right)$ is a perfect square, and when taking “+”, $m=94 n-21$; when taking “-”, $m=94 n-25$.

After calculation, the smallest positive integer $n$ that makes $47 n^{2}-2 n+1$ a perfect square is $n=6$, at which time, $m=94 \times 6-25=539$; $47 n^{2}+2 n+1$ is a perfect square for the smallest positive integer $n=8$, at which time, $m=94 \times 8-21=731$.

Therefore, the smallest two positive integers $m$ that make $47\left(m^{2}+46 m+713\right)$ a perfect square are 539 and 731.