7 (1) Find the number of positive integer solutions $(m, n, r)$ to the indeterminate equation $m n+n r+m r=2(m+n+r)$;
(2) For a given integer $k>1$, prove that the indeterminate equation $m n+n r+m r=k(m+n+r)$ has at least $3 k+1$ sets of positive integer solutions $(m, n, r)$.
(1) If $m, n, r \geqslant 2$, then
$$
m n \geqslant 2 m, n r \geqslant 2 n, m r \geqslant 2 r
$$

Thus,
$$
m n+n r+m r \geqslant 2(m+n+r),
$$

Therefore, the above inequalities all hold with equality, so $m=n=r=2$.
If $1 \in\{m, n, r\}$, assume without loss of generality that $m=1$, then $n r+n+r=2(1+n+r)$, hence $(n-1)(r-1)=3$, so $\{n-1, r-1\}=\{1,3\}$, thus $\{n, r\}=\{2,4\},\{m, n, r\}=\{1,2,4\}$, and there are $3!=6$ such solutions.

Therefore, the indeterminate equation $m n+n r+m r=2(m+n+r)$ has a total of 7 sets of positive integer solutions.
(2) Transform $m n+n r+m r=k(m+n+r)$ into
$$
\begin{array}{c}
{[n-(k-m)][r-(k-m)]=k^{2}-k m+m^{2} .} \\
n=k-m+1, r=k^{2}-k m+m^{2}+k-m
\end{array}
$$

This satisfies the equation. When $m=1,2, \cdots,\left[\frac{k}{2}\right]$, $0<m<n<r$.
When $k$ is even,
$$
\{m, n, r\}=\left\{l, k-l+1, k^{2}-k l+l^{2}+k-l\right\},
$$

where $l=1,2, \cdots, \frac{k}{2}$ provides $3 k$ sets of positive integer solutions for the indeterminate equation.
When $k$ is odd,
$$
\{m, n, r\}=\left\{l, k-l+1, k^{2}-k l+l^{2}+k-l\right\},
$$

where $l=1,2, \cdots, \frac{k-1}{2}$ provides $3(k-1)$ sets of positive integer solutions for the indeterminate equation, and there are two $\frac{k+1}{2}$ and one
$$
k^{2}-k \frac{k+1}{2}+\left(\frac{k+1}{2}\right)^{2}+k-\frac{k+1}{2}=\frac{(k+1)(3 k-1)}{4}
$$

cases provide 3 sets of positive integer solutions for the indeterminate equation.
And $m=n=r=k$ is also a positive integer solution for the indeterminate equation.
Therefore, the indeterminate equation $m n+n r+m r=k(m+n+r)$ has at least $3 k+1$ sets of positive integer solutions.