Find all integers $n \geq 1$ such that $2^{n}-1$ has exactly $n$ positive integer divisors.
The valid $n$ are $1,2,4,6,8,16,32$. They can be verified to work through inspection, using the well known fact that the Fermat prime $F_{i}=2^{2^{i}}+1$ is indeed prime for $i=0,1, \ldots, 4$ (but not prime when $i=5$ ).

We turn to the proof that these are the only valid values of $n$. In both solutions that follow, $d(n)$ is the divisor counting function.

『 First approach (from author). Let $d$ be the divisor count function. Now suppose $n$ works, and write $n=2^{k} m$ with $m$ odd. Observe that
$$
2^{n}-1=\left(2^{m}-1\right)\left(2^{m}+1\right)\left(2^{2 m}+1\right) \cdots\left(2^{2^{k-1} m}+1\right),
$$
and all $k+1$ factors on the RHS are pairwise coprime. In particular,
$$
d\left(2^{m}-1\right) d\left(2^{m}+1\right) d\left(2^{2 m}+1\right) \cdots d\left(2^{2^{k-1} m}+1\right)=2^{k} m .
$$

Recall the following fact, which follows from Mihǎilescu's theorem.
Lemma
$2^{r}-1$ is a square if and only if $r=1$, and $2^{r}+1$ is a square if and only if $r=3$.
Now, if $m \geq 5$, then all $k+1$ factors on the LHS are even, a contradiction. Thus $m \leq 3$. We deal with both cases.

If $m=1$, then the inequalities
$$
\begin{array}{c}
d\left(2^{2^{0}}-1\right)=1 \\
d\left(2^{2^{0}}+1\right) \geq 2 \\
d\left(2^{2^{1}}+1\right) \geq 2 \\
\vdots \\
d\left(2^{2^{k-1}}+1\right) \geq 2
\end{array}
$$
mean that it is necessary and sufficient for all of $2^{2^{0}}+1,2^{2^{1}}+1, \ldots, 2^{2^{k-1}}+1$ to be prime. As mentioned at the start of the problem, this happens if and only if $k \leq 5$, giving the answers $n \in\{1,2,4,8,16,32\}$.

If $m=3$, then the inequalities
$$
\begin{array}{c}
d\left(2^{3 \cdot 2^{0}}-1\right)=2 \\
d\left(2^{3 \cdot 2^{0}}+1\right)=3 \\
d\left(2^{3 \cdot 2^{1}}+1\right) \geq 4 \\
\vdots \\
d\left(2^{3 \cdot 2^{k-1}}+1\right) \geq 4
\end{array}
$$
mean that $k \geq 2$ does not lead to a solution. Thus $k \leq 1$, and the only valid possibility turns out to be $n=6$.
Consolidating both cases, we obtain the claimed answer $n \in\{1,2,4,6,8,16,32\}$.
₪ Second approach using Zsigmondy (suggested by reviewers). There are several variations of this Zsigmondy solution; we present the approach found by Nikolai Beluhov. Assume $n \geq 7$, and let $n=\prod_{1}^{m} p_{i}^{e_{i}}$ be the prime factorization with $e_{i}>0$ for each $i$. Define the numbers
$$
\begin{aligned}
T_{1} & =2^{p_{1}^{c_{1}}}-1 \\
T_{2} & =2^{p_{2}^{c_{2}}}-1 \\
& \vdots \\
T_{m} & =2^{p_{m}^{c_{m}}}-1 .
\end{aligned}
$$

We are going to use two facts about $T_{i}$.
Claim - The $T_{i}$ are pairwise relatively prime and
$$
\prod_{i=1}^{m} T_{1} \mid 2^{n}-1 .
$$

Proof. Each $T_{i}$ divides $2^{n}-1$, and the relatively prime part follows from the identity $\operatorname{gcd}\left(2^{x}-1,2^{y}-1\right)=2^{\operatorname{gcd}(x, y)}-1$.

Claim - The number $T_{i}$ has at least $e_{i}$ distinct prime factors.
Proof. This follows from Zsigmondy's theorem: each successive quotient $\left(2^{p^{k+1}}-1\right) /\left(2^{p^{k}}-\right.$ 1) has a new prime factor.

Claim (Main claim) - Assume $n$ satisfies the problem conditions. Then both the previous claims are sharp in the following sense: each $T_{i}$ has exactly $e_{i}$ distinct prime divisors, and
$$
\left\{\text { primes dividing } \prod_{i=1}^{m} T_{i}\right\}=\left\{\text { primes dividing } 2^{n}-1\right\} .
$$

Proof. Rather than try to give a size contradiction directly from here, the idea is to define an ancillary function
$$
s(x)=\sum_{p \text { prime }} \nu_{p}(x)
$$
which computes the sum of the exponents in the prime factorization. For example
$$
s(n)=e_{1}+e_{2}+\cdots+e_{m} .
$$

On the other hand, using the earlier claim, we get
$$
s\left(d\left(2^{n}-1\right)\right) \geq s\left(d\left(\prod T_{i}\right)\right) \geq e_{1}+e_{2}+\cdots+e_{m}=s(n) .
$$

But we were told that $d\left(2^{n}-1\right)=n$; hence equality holds in all our estimates, as needed.

At this point, we may conclude directly that $m=1$ in any solution; indeed if $m \geq 2$ and $n \geq 7$, Zsigmondy's theorem promises a primitive prime divisor of $2^{n}-1$ not dividing any of the $T_{i}$.
Now suppose $n=p^{e}$, and $d\left(2^{p^{e}}-1\right)=n=p^{e}$. Since $2^{p^{e}}-1$ has exactly $e$ distinct prime divisors, this can only happen if in fact
$$
2^{p^{e}}-1=q_{1}^{p-1} q_{2}^{p-1} \ldots q_{e}^{p-1}
$$
for some distinct primes $q_{1}, q_{2}, \ldots, q_{e}$. This is impossible modulo 4 unless $p=2$. So we are left with just the case $n=2^{e}$, and need to prove $e \leq 5$. The proof consists of simply remarking that $2^{2^{5}}+1$ is known to not be prime, and hence for $e \geq 6$ the number $2^{2^{e}}-1$ always has at least $e+1$ distinct prime factors.