Finite Number of Principal Ideals 
Let $R$ be a unique factorization domain and $d$ a nonzero element of $R$. There are only a finite number of distinct principal ideals that contain the ideal $(d)$. 

Suppose that $d \neq 0$ is not a unit. Then $d = d_1 ... d_n$ for irreducible/prime elements $d_i$; and suppose that $(d) \subseteq (k)$, where $k \neq 0$ is not a unit either, which means $k=k_1...k_m$ for irreducibles/primes $k_i$. Then $d = kx$ for some $x \neq 0$. If $x$ is a unit, then $dx^{-1} =k$ and therefore $(d)=(k)$, which is very uninteresting. So suppose that $x$ is not a unit, and therefore $x=x_1...x_s$ for irreducibles/primes $x_i$. Then $d=kx$ becomes $d_1...d_n = k_1...k_m x_1 ... x_s$, and $R$ being a UFD entails that $n = m+s$, where $n \in \Bbb{N}$ is fixed.
So, for each $k \in R$ such that $(d) \subseteq (k)$, there exist integers $m,s \in \Bbb{N}$ such that $n=m+s$, and so to count the total number of $k$ satisfying $(d) \subseteq (k)$ is to count a number that is bounded by the total number of ways of writing a fixed positve integer $n$ as the sum of two positive integers, which is finite. Hence, the number of $k$ with $(d) \subseteq (k)$ is finite, and in fact bounded below by $n$ since $(d) \subseteq (d_i)$ for $i=1,...,n$.
How does this sound?  
EDIT: 
Here is another attempt, although I wasn't exactly able to do what leibnewtz suggested. Starting with $d_1...d_n = k_1...k_nx_1...x_s$, where every element involved in the equation is a prime/irreducible element, we see that every $k_i$ must be an associate of some $d_i$. WLOG, suppose that $k_i$ and $d_i$ are associates, and therefore $k_i = u_i d_i$ for some unit $u_i$. Then $k = k_1 ... k_m = u_1 d_1 .... u_n d_m = (u_1....u_n)d_1...d_m = u d_1 ... d_m$, where $u = u_1...u_m$ is a unit, which means $(k) = (u d_1 ... d_m) = (d_1 ... d_m)$. Thus, for each $k$ satisfying $(d) \subseteq (k)$, $(k)$ is just the ideal generated by some finite product of the $d_i$ with nonrepeating factors, of which there are only finitely many. 
How does this sound? 
 A: As per leibnewt's suggestion, here is what I believe is a solution to the problem:
Suppose that $d \neq 0$ is not a unit. Then $d = d_1 ... d_n$ for irreducible/prime elements $d_i$; and suppose that $(d) \subseteq (k)$, where $k \neq 0$ is not a unit either, which means $k=k_1...k_m$ for irreducibles/primes $k_i$. Then $d = kx$ for some $x \neq 0$. If $x$ is a unit, then $dx^{-1} =k$ and therefore $(d)=(k)$, which is very uninteresting. So suppose that $x$ is not a unit, and therefore $x=x_1...x_s$ for irreducibles/primes $x_i$. Then we have $d_1...d_n = k_1...k_nx_1...x_s$, where every element involved in the equation is a prime/irreducible element, we see that every $k_i$ must be an associate of some $d_i$. WLOG, suppose that $k_i$ and $d_i$ are associates, and therefore $k_i = u_i d_i$ for some unit $u_i$. Then $k = k_1 ... k_m = u_1 d_1 .... u_n d_m = (u_1....u_n)d_1...d_m = u d_1 ... d_m$, where $u = u_1...u_m$ is a unit, which means $(k) = (u d_1 ... d_m) = (d_1 ... d_m)$. Thus, for each $k$ satisfying $(d) \subseteq (k)$, $(k)$ is just the ideal generated by some finite product of the $d_i$ with nonrepeating factors, of which there are only finitely many. 
