* 6 . If $2^{10}+2^{13}+2^{m}$ is a perfect square, then the natural number $m=$ $\qquad$
6. $m=14$. Let $k^{2}=2^{10}+2^{13}+2^{m}$, then
$$
\begin{aligned}
2^{m} & =k^{2}-\left(2^{10}+2^{13}\right)=k^{2}-\left(2^{5} \cdot 3\right)^{2} \\
& =(k+96)(k-96) .
\end{aligned}
$$

Therefore, there exist two non-negative integers $s$ and $t$, such that
$$
\left\{\begin{array}{l}
k-96=2^{s}, \\
k+96=2^{m-t}=2^{t}, s<t .
\end{array}\right.
$$

Eliminating $k$ yields $2^{s}\left(2^{t-s}-1\right)=2^{6} \cdot 3$. Since $2^{t-s}-1$ is odd, we get $s=6,2^{t-s}-1=3$. Thus, $t=8, m=s+t=14$.