30. Find all integers $n > 1$ such that any of its divisors greater than 1 have the form $a^{r}+1$, where $a \in \mathbf{N}^{*}, r \geqslant 2, r \in \mathbf{N}^{*}$.
30. Let all numbers that meet the conditions form a set $S$. For $x \in S(x>2)$, it is known that there exist $a, r \in \mathbf{N}^{\cdot}$, such that $x=a^{r}+1$, where $a, r \geqslant 2$. For this $x$, we set $a$ to be the smallest positive integer in this representation of $x$, then $r$ must be even (otherwise $(a+1) \mid x$, thus there exist $b, t \in \mathbf{N}^{*}, t \geqslant 2$, such that $a+1=b^{t}+1$, i.e., $a=b^{t}$, which leads to $x=b^{*}+1$ contradicting the minimality of $a$). Therefore, every number in $S$ can be expressed in the form $m^{2}+1$ (note that $2 \in S$, and also $2=1^{2}+1$), where $m \in \mathbf{N}^{*}$.

Below, we use the above property to determine the elements $p$ in $S$.
If $p$ is a prime, then $p$ is a prime of the form $m^{2}+1\left(m \in \mathbf{N}^{*}\right)$.
If $p$ is an odd composite number, then there exist odd primes $u \leqslant v$, such that $u, v, u v \in S$, thus there exist $a, b, c \in \mathbf{N}^{*}$, such that $a \leqslant b<c$, and
$$u=4 a^{2}+1, v=4 b^{2}+1, u v=4 c^{2}+1$$

At this time, $v(u-1)=4\left(c^{2}-b^{2}\right)$, hence $v \mid(c-b)$ or $v \mid(c+b)$, both leading to $v<2 c, u v \leqslant v^{2}<4 c^{2}$, a contradiction. Therefore, $p$ is not an odd composite number.

If $p$ is an even composite number, note that $4 \notin S$, and odd composites $\notin S$, so we can set $p=2 q$, where $q$ is an odd prime. In this case, there exist $a, b \in \mathbf{N}^{*}$, such that
$$q=4 a^{2}+1,2 q=b^{2}+1$$

Thus, $q=b^{2}-4 a^{2}=(b-2 a)(b+2 a)$, so
$$(b-2 a, b+2 a)=(1, q)$$

This gives $q=4 a+1$, so $4 a^{2}+1=4 a+1$, yielding $a=1, q=5$, and thus $p=10$.
In summary, $S=\left\{10\right.$ or primes of the form $a^{2}+1$, where $\left.a \in \mathbf{N}^{*}\right\}$ (note that the numbers described in the set clearly meet the requirements).