11. Let $n$ be a positive integer, and $n^{2}+1085$ is a positive integer power of 3. Then the value of $n$ is $\qquad$
11.74.

Generally, the unit digit of $n^{2}$ (where $n$ is a positive integer) can only be 0, 1, 4, 5, 6, 9, so the unit digit of $n^{2}+1085$ can only be 5, 6, 9, 0, 1, 4; while the unit digit of $3^{m}$ (where $m$ is a positive integer) can only be 1, 3, 7, 9.

From the given, let $n^{2}+1085=3^{m}$ (where $n, m$ are positive integers), we can deduce that the unit digit of $3^{m}$ can only be 1 or 9, and $m$ is an even number.
Let $m=2k$ (where $k$ is a positive integer), then we have
$$
n^{2}+1085=3^{2k} \text{. }
$$

Rearranging gives $\left(3^{k}-n\right)\left(3^{k}+n\right)=1 \times 5 \times 7 \times 31$.
We get $\left\{\begin{array}{l}3^{k}-n=1, \\ 3^{k}+n=1085 ;\end{array}\left\{\begin{array}{l}3^{k}-n=5, \\ 3^{k}+n=217 ;\end{array}\right.\right.$
$$
\left\{\begin{array}{l}
3^{k}-n=7, \\
3^{k}+n=155 ;
\end{array} \left\{\begin{array}{l}
3^{k}-n=31, \\
3^{k}+n=35 .
\end{array}\right.\right.
$$

However, only the equation set $\left\{\begin{array}{l}3^{k}-n=7, \\ 3^{k}+n=155\end{array}\right.$ has a solution that meets the conditions: $\left\{\begin{array}{l}k=4, \\ n=74 .\end{array}\right.$