Actually, no. In any valuation domain, the prime ideals are linearly ordered by inclusion, so there exists at most one nonzero principal prime ideal.
In particular, let $K$ be any field, and let $$R=K[x,y/x,y/x^2,y/x^3,...],$$
i.e., elements of $R$ are "polynomials" in $x$ and $y$ over $K$, except you can divide $y$ by $x$ as many times as you like.
Consider the subset $S$ of $R$ containing all elements with nonzero constant term. It is clear that $S$ is a multiplicative subset of $R$, and let $T=R_S$ be the localization of $R$ at $S$--i.e., $$T=\left\{\frac{f(x,y)}{g(x,y)}\,\vert\,f(x,y)\in R, g(x,y)\in S\right\}.$$
It isn't difficult to show that any nonzero element of $T$ is of the form $ux^n y^m$ where $u\in U(T)$, $m\geq 0$, and if $m=0$ then $n\geq 0$. (Basically, take an arbitrary nonzero element of $T$ and factor out all of the $x$'s and $y$'s that you can.)
So, we have the following chain of prime ideals in $T$ (and, in fact, these are all the prime ideals of $T$): $$0\subsetneq (y,y/x,y/x^2,y/x^3,\cdots)\subsetneq xT$$
You can generalize this to a chain of length $n$ by taking a valuation domain with value group isomorphic to $\mathbb{Z}^n$ under the lexicographic ordering.
Your idea that finitely generated ideals are "small" and non-finitely generated ideals are "large" is misleading. Think about $R = \Bbb{Z}[X_1, X_2, \ldots]$, the ring of polynomials over the integers in countably many variables $X_1, X_2, \ldots$ If you define $J_k = \langle X_1, \ldots, X_k\rangle$, then $J_1 \subseteq J_2 \subseteq \ldots$ is a strictly increasing chain of ideals, so $R$ is not Noetherian. The finitely generated ideal $J = \langle 2 \rangle $ properly contains the non-finitely generated ideal $K = \langle 2X_1, 2X_2, \ldots \rangle$. For any prime $p$, the ideal $L_p$ comprising the polynomials with constant term a multiple of $p$ is maximal non-finitely generated.
Best Answer
In $\mathbb{Z}[X]$, the ideal generated by $2X$ and $3X$ is $(X)$ because $X = 3X-2X$, so it is prime.
However, $2X$ is not prime because $2X$ divides $2\times X$, but it does not divide $2$ nor $X$ in $\mathbb{Z}[X]$. The same goes for $3X$.