7. For each positive integer $n \leqslant m$, we have
$$
\frac{2 n+1}{3 n+8}<\frac{\sqrt{5}-1}{2}<\frac{n+7}{2 n+1} \text {. }
$$

Then the positive integer $m=$
7. 27

From the original inequality, we get
$$
\begin{array}{l}
\frac{7 n+10}{3 n+8}<\sqrt{5}<\frac{4 n+15}{2 n+1} \\
\Rightarrow \frac{49 n^{2}+140 n+100}{9 n^{2}+48 n+64}<5 \\
\quad<\frac{16 n^{2}+120 n+225}{4 n^{2}+4 n+1} \\
\Rightarrow n^{2}-25 n-55<0 \\
\Rightarrow n(n-25)<55 \\
\Rightarrow n(n-25) \leqslant 54 \Rightarrow n \leqslant 27 .
\end{array}
$$