Take $n^4+1 = 2^{\displaystyle{ 4 \cdot 3^{1000} }}+1 $. Clearly $$2^{\displaystyle{ 4 \cdot 3^{0} }}+1 , 2^{\displaystyle{ 4 \cdot 3^{1} }}+1 , \ldots , 2^{\displaystyle{ 4 \cdot 3^{1000} }}+1 $$Are all divisors of $2^{\displaystyle{ 4 \cdot 3^{1000} }}+1 $ since $2^a+1 \mid 2^b+1 $ for every $ \frac{b}{a} $ odd. Proven

My solution We will take a sequence $a_1,a_2,\ldots, a_{1001} $ such that $a_1=1$ and for every $n$, $$a_{n+1} = (a_1^4+1)(a_2^4+1)\ldots(a_n^4+1).$$So we have $\text{GCD}(a_i^4+1, a_j^4+1)=1 $ for every $i\ne j$. We will take $n$ such that \begin{align*} n &\equiv a_1 \pmod {a_1^4+1} &\implies a_1^4+1 \mid n^4+1 \\ n &\equiv a_2 \pmod {a_2^4+1} &\implies a_2^4+1 \mid n^4+1 \\ &\vdots \\ n &\equiv a_{1001} \pmod {a_{1001}^4+1} &\implies a_{1001}^4+1 \mid n^4+1 \end{align*}By CRT, there exist such $n$. Proven.

Call such a number of the form $n^4+1$ fancy. It is sufficient to show that every fancy number has a strictly larger fancy multiple so we can simply notice that $$n^4+1|(n^4+n+1)^4+1$$

Construct the sequence $a_1, a_2, \cdots$ as follows: $$a_1 = 1, a_{n+1} = (a_n^4 + 2)a_n \forall n \ge 1.$$Then $a_i^4 +1 \mid a_{i+1}^4 +1, a_{i+1} > a_i$. Therefore, $a_{1001}^4+1$ has at least 1001 divisors of the given form. $\square$

AshAuktober wrote: Construct the sequence $a_1, a_2, \cdots$ as follows: $$a_1 = 1, a_{n+1} = (a_n^4 + 2)a_n \forall n \ge 1.$$Then $a_i^4 +1 \mid a_{i+1}^4 +1, a_{i+1} > a_i$. Therefore, $a_{1001}$ has at least 1001 divisors of the given form. $\square$ Why is $a_{1001}$ of the given $n^4 + 1$ form?

RevolveWithMe101 wrote: AshAuktober wrote: Construct the sequence $a_1, a_2, \cdots$ as follows: $$a_1 = 1, a_{n+1} = (a_n^4 + 2)a_n \forall n \ge 1.$$Then $a_i^4 +1 \mid a_{i+1}^4 +1, a_{i+1} > a_i$. Therefore, $a_{1001}$ has at least 1001 divisors of the given form. $\square$ Why is $a_{1001}$ of the given $n^4 + 1$ form? Whoops, typo, fixed

Sol by MATHEMATICAL717:- Consider $n^4+1=N=2^{3^{1000} * 4}+1$. We know all numbers of form $2^{3^a * 4}+1$ where $a \in \{0,1,...,1000\}$ are divisors of $N$ since $\frac{3^{1000} * 4}{3^a * 4}=3^{1000-a}$ is an odd number. Thus we are done.

Recall that for any polynomial $P(n) \in \mathbb Z[x]$ we have $a-b \mid P(a)-P(b)$ therefore $P(n)+n-n \mid P(P(n)+n)-P(n)$ which means that $P(n) \mid P(P(n)+n)$, set $P(x)=x^4+1$ here to get that $n^4+1 \mid (n^4+n+1)^4+1$, therefore if we consider the sequence of polynomials defined as $P_1(x)=x^4+1$ and for $i \in \mathbb Z_{>0}$, $P_{i+1}(x)=P_i(P_i(x)+x)$ then by considering $P_{1001}(n)$ we are easly done.