(Since there is no answer yet, I removed most "EDIT"'s to make the text more readable)
Today, I was trying to find a natural number $n_{2}$ such that this number has the same distinct prime factors as another, previous number $n_{1}$ with $n_{2}>n_{1}$, but does not differ too much from it. The extreme would be a difference of $2$ because $gcd(n,n+1)=1$.
There is one trivial example: $n_{1}=2$ and $n_{2}=2+2=4=2^2$. Another example (which is wrong, but illustrates the problem) is something like: $n_{1}=2^{80}*3^{90}*5^{100}$ and $n_{2}=n_{1}+2=2^{82}*3^{89}*5^{100}$. Notice how the exponents can be smaller than in the previous number.
In my case composite numbers are more interesting. It would actually be nice if there was an efficient algorithm for this (I am aware that prime factorization is a hard problem in general, which could be important here I assume).
Maybe we should assume that the distribution of these primes where this condition is met is crucial for a fast algorithm because we could just search by incrementing $n_{1}$.
To phrase this as a search problem:
Input: $k,d\in \mathbb{N}$.
Problem: Find a (small) number $n_{1}=p_{1}^{e_{1}^{(1)}}p_{2}^{e_{2}^{(1)}}...p_{k}^{e_{k}^{(1)}}$ with exponents $e_{1}^{(1)},...,e_{k}^{(1)}>0$ such that $n_{2}=n_{1}+d$ has the same $k$ distinct prime factors with exponents $e_{1}^{(2)},...,e_{k}^{(2)}>0$.
Suggestions for finding a solution to $d=2$ would be awesome. Perhaps this problem is already well known.
EDIT: Alright, from the comments:
It seems like $d$ might not be "enough" at some point (when $k$ grows), meaning that the number of cases where the conditions are met for a specific $d$ is always finite and therefore it might be wise to not let the problem input depend on $d$, but soley on $k$. Maybe this is not true. Since this would help refining the problem statement, maybe someone knows how to deal with this behaviour.
The new problem would be:
Input: $k\in \mathbb{N}$.
Problem: Find the least $d$ and a (small) number $n_{1}=p_{1}^{e_{1}^{(1)}}p_{2}^{e_{2}^{(1)}}...p_{k}^{e_{k}^{(1)}}$ with $e_{1}^{(1)},...,e_{k}^{(1)}>0$ such that $n_{2}=n_{1}+d$ has the same $k$ distinct prime factors with $e_{1}^{(2)},...,e_{k}^{(2)}>0$.
(Okay, actually it would not be a huge problem if some exponents were equal to zero for $n_{2}$. But then I would still need to be able to have some lower bound for the number of distinct prime factors that does not grow too slowly with $k$.)
The updated problem is more general, so let's consider that one.
Note that $n_{2} = n_{1} + d$ implies that $gcd(n1, n2) \vert d$. Since $n_{1}=p_{1}^{e_{1}^{(1)}}p_{2}^{e_{2}^{(1)}}...p_{k}^{e_{k}^{(1)}}$ and $n_{2}=p_{1}^{e_{1}^{(2)}}p_{2}^{e_{2}^{(2)}}...p_{k}^{e_{k}^{(2)}}$, and for each $i$ and $j$, $e_{i}^{({j})} > 0$, then $gcd(n1, n2) >= p_{1}p_{2}...p_{k}$. For a fixed $k$, this means that the least $d$ that satisfies the conditions will be the smallest positive integer that has $k$ positive factors, i.e. $d=\Pi_{i=1}^k P_{i}$, where $P_{i}$ is the $i$-th prime number. The least $n1$ and $n2$ that satisfy the problem will be $n_{1}=d$ and $n_{2} = 2d$.
This result can also be applied to the original problem. For any fixed $k$ and $d = p_{1}^{e_{1}}p_{2}^{e_{2}}...p_{l}^{e_{l}}$, it follows that there will only be solutions if $k <= l$. For the special case $d=2$, i.e. $l = 1$, there will only be a single solution (the one you already mentioned in your question).
There's also a question of whether, whenever there is a solution for a fixed $k$ and $d$, there are infinite solutions as well. For $k = 2$, this problem can be reduced to solving the Catalan's Conjecture, which is now known to be true, i.e. there is only a single solution for $6 \nmid d$, and only two solutions $d = 2 \cdot 3$. For $k > 3$, the problem seems to be a special case of Pillai's conjecture for $C = 1$, and $A, B, x, y$ having all of their divisors intersect with those of $d$.