15. Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1, a_{n+1}=\frac{1}{8} a_{n}^{2}+m\left(n \in \mathbf{N}^{*}\right)$, if for any positive integer $n$, we have $a_{n}<4$, find the maximum value of the real number $m$.
15. Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1, a_{n+1}=\frac{1}{8} a_{n}^{2}+m\left(n \in \mathbf{N}^{*}\right)$, if for any positive integer $n$, we have $a_{n}<4$, note that as $n \rightarrow+\infty$, $(m-2)(n-1) \rightarrow+\infty$,
Therefore, there exists a sufficiently large $n$ such that $1+(m-2)(n-1)>4$, i.e., $a_{n}>4$, which is a contradiction!

Thus, $m \leq 2$. $\qquad$
Furthermore, when $m=2$, it can be proven that for any positive integer $n$, we have $0<a_{n}<4$.
When $n=1, a_{1}=1<4$, the conclusion holds;
Assume that for $n=k(k \geq 1)$, the conclusion holds, i.e., $0<a_{k}<4$,
Then $0<a_{k+1}=2+\frac{1}{8} a_{k}^{2}<2+\frac{1}{8} \times 4^{2}=4$,
i.e., the conclusion holds for $n=k+1$.
By the principle of mathematical induction, for any positive integer $n$, we have $0<a_{n}<4$.
In summary, the maximum value of the real number $m$ is 2.
20 points