15. Find all positive integers $n$, such that $n+36$ is a perfect square, and apart from 2 or 3, $n$ has no other prime factors.
15. Let $n+36=(x+6)^{2}\left(x \in \mathbf{N}_{+}\right)$, then $n=x(x+12)$.
According to the problem, we can set $\left\{\begin{array}{l}x=2^{a_{1}} \times 3^{b_{1}}, \\ x+12=2^{a_{2}} \times 3^{b_{2}},\end{array}\right.$ where $a_{1}, a_{2}, b_{1}, b_{2}$ are all non-negative integers. Thus,
$$
2^{a_{2}} \times 3^{b_{2}}-2^{a_{1}} \times 3^{b_{1}}=12.
$$

If $a_{1}=a_{2}=0$, then $3^{b_{2}}-3^{b_{1}}=12$, which is impossible. Therefore, at least one of $a_{1}, a_{2}$ is greater than 0. Thus, both $x$ and $x+12$ are even. Hence, $a_{1}, a_{2}$ are both positive integers.
If $a_{2}=1$, then $2 \times 3^{b_{2}}=12+2^{a_{1}} \times 3^{b_{1}}$.
Clearly, only $a_{1}=1$ is possible (otherwise, the remainders when divided by 4 on both sides are different), at this time, $3^{b_{2}}=6+3^{b_{1}}$.
Clearly, only $b_{2}=2, b_{1}=1$ is possible, at this time,
$$
x=6, n=108.
$$

If $a_{2} \geqslant 2$, then $x+12$ is a multiple of 4. Thus, $x$ is also a multiple of 4, so $a_{1} \geqslant 2$. At this time,
$$
2^{a_{2}-2} \times 3^{b_{2}}-2^{a_{1}-2} \times 3^{b_{1}}=3.
$$

Clearly, at least one of $a_{1}-2, a_{2}-2$ should be 0 (otherwise, the parity of both sides of equation (1) is different).
(1) When $a_{2}-2=0$, i.e., $a_{2}=2$,
$$
3^{b_{2}}-2^{a_{1}-2} \times 3^{b_{1}}=3.
$$

At this time, $a_{1}-2>0$ (otherwise, the parity of both sides of equation (2) is different), so $b_{2}>b_{1}$.

If $b_{1} \geqslant 2$, then the left side of equation (2) is a multiple of 9, while the right side is 3, which is a contradiction. Therefore, only $b_{1}=1$. Thus, equation (2) becomes $3^{b_{2}-1}-2^{a_{1}-2}=1$, which has only two solutions
$$
\left\{\begin{array}{l}
a_{1}-2=1, \\
b_{2}-1=1
\end{array}\right. \text{ and } \left\{\begin{array}{l}
a_{1}-2=3, \\
b_{2}-1=2,
\end{array}\right.
$$

i.e., $\left\{\begin{array}{l}a_{1}=3, \\ b_{2}=2\end{array}\right.$ and $\left\{\begin{array}{l}a_{1}=5, \\ b_{2}=3 .\end{array}\right.$
At this time, the corresponding $x$ values are 24 and 96, and the corresponding $n$ values are 864 and 10368.
(2) When $a_{1}-2=0$, i.e., $a_{1}=2$,
$$
2^{a_{2}-2} \times 3^{b_{2}}-3^{b_{1}}=3.
$$

At this time, $a_{2}-2>0$ (otherwise, the parity of both sides of equation (3) is different), so $b_{2} \leqslant b_{1}$.

If $b_{2} \geqslant 2$, then the left side of equation (3) is a multiple of 9, while the right side is 3, which is a contradiction. Therefore, $b_{2} \leqslant 1$.

If $b_{2}=0$, then $2^{a_{2}-2}-3^{b_{1}}=3$, only $b_{1}=0$ is possible, at this time, $a_{2}=4, x=4, n=64$.

If $b_{2}=1$, then $2^{a_{2}-2}-3^{b_{1}-1}=1$, which has only two solutions
$$
\left\{\begin{array}{l}
a_{2}-2=1, \\
b_{1}-1=0
\end{array}\right. \text{ and } \left\{\begin{array}{l}
a_{2}-2=2, \\
b_{1}-1=1,
\end{array}\right.
$$

i.e., $\left\{\begin{array}{l}a_{2}=3, \\ b_{1}=1\end{array}\right.$ and $\left\{\begin{array}{l}a_{2}=4, \\ b_{1}=2 .\end{array}\right.$
At this time, the corresponding $x$ values are 12 and 36, and the corresponding $n$ values are 288 and 1728.
Therefore, there are 6 values of $n$ that meet the conditions, which are
$$
64,108,288,864,1728,10368.
$$
(Provided by Xu Shenglin)