12 Find all positive integers $n>1$, such that any of its positive divisors greater than 1 has the form $a^{r}+1$, where $a$ is a positive integer, and $r$ is a positive integer greater than 1.
Let $n$ be a positive integer that meets the condition, then every divisor of $n$ greater than 1 also meets the condition.

When $n>2$, we can write $n=a^{r}+1(r>1, a>1)$, such that $a$ is the smallest value, then $r$ is even. Because if $r$ is odd, then $a+1 \mid n$, so there exist positive integers $b, t(t>1)$, such that $a+1=b^{t}+1$, which gives $a=b^{t}, n=b^{r t}+1$, contradicting the minimality of $a$. Therefore, $n$ equals a perfect square plus 1.

Obviously, primes of the form $a^{2}+1$ meet the condition. Now consider the case where $n$ is a composite number.

If $n$ has two odd prime factors $p, q(p, q$ can be the same), then $p, q$ and $p q$ all meet the condition, i.e., there exist positive integers $a, b, c$, such that $p=4 a^{2}+1, q=4 b^{2}+1$, $p q=4 c^{2}+1$, thus
$$
\left(4 a^{2}+1\right)\left(4 b^{2}+1\right)=4 c^{2}+1 .
$$

Assume $a \leqslant b$, then from (1) we have
$$
4 a^{2}\left(4 b^{2}+1\right)=4(c-b)(c+b) .
$$

Since $4 b^{2}+1$ is a prime, we know that $4 b^{2}+1 \mid c-b$ or $4 b^{2}+1 \mid c+b$, hence $c \geqslant 4 b^{2}-b+1$. Using $a \leqslant b$, substituting into (1), it is easy to see that the right side of (1) is larger than the left side. Contradiction.

Since 4 is not a number that meets the condition, the composite number $n=2 p, p$ is an odd prime. In this case, there exist positive integers $a, b$, such that
$$
2\left(a^{2}+1\right)=b^{2}+1 .
$$

That is, $a^{2}+1=(b-a)(b+a)$, since $a^{2}+1$ is a prime, so, $(b-a, b+a)=\left(1, a^{2}+1\right)$, solving gives $a=2, b=3$, hence $n=10$.
In summary, the numbers that meet the condition are 10 or primes of the form $a^{2}+1$.