I have been thinking about this question from an algebra qual exam for some time but have not been able to figure it out. I feel like there should be a very simple example of this in a polynomial ring such as $\mathbb{Z}[x]$. But I have neither found an example nor disproved it.
I considered ideals of the form $I = (n, f(x))\subset\mathbb{Z}[x]$, where $n$ is composite number but have not made much progress from there.
Work in $k[x,y]$ and multiply $$(x) \subset (x,y) \subset (1)$$ with $(x)$.
Or, since you wanted $\mathbb Z[x]$: Multiply $$(2) \subset (2,x) \subset (1)$$ with $(2)$.
For the sake of fun, let me do it the most abstract way I could come up with: Let $R$ be an UFD, which is not a PID and $p$ be an irreducible element.
There is no principal ideal properly between $(p^2)$ and $(p)$. But there is certainly at least one ideal properly between $(p^2)$ and $(p)$, because the set of ideals between $(p^2)$ and $(p)$ corresponds to the set of submodules of $(p)/(p^2) \cong R/(p)$. The latter has a proper non-zero ideal, because it is not a field. Thus we have proven the existence of a non-principal ideal with radical equal to $(p)$.