$\mathrm{Spec}(R)\!=\!\mathrm{Max}(R)\!\cup\!\{0\}$ $\Rightarrow$ $R$ is a PID Is the following true:

If $R$ is a commutative unital ring with
  $\mathrm{Spec}(R)\!=\!\mathrm{Max}(R)\!\cup\!\{0\}$, then $R$ is a PID.

If yes, how can one prove it?
Since $0$ is a prime ideal, $R$ is a domain. Thus we must prove that every ideal is principal. I'm not sure if this link (first answer) helps.
 A: Sanchez's example is from algebraic geometry. Here's an example from algebraic number theory.
The ring of integers of an algebraic number field is always a Dedekind domain; in particular, it is a 1-dimensional domain. However, there are (plenty of) ring of integers which are not PIDs. As you can see here, only a finite number of imaginary quadratic extensions of $\mathbb{Q}$ have a PID as ring of integers.
A: You are claiming that every 1-dimension domain is PID, which is not true. For example, take a curve which is not smooth (Say $\mathbb{C}[x,y]/(y^2−x^3)$) will give you a counterexample, since it won't be a PID when you localize at $(x,y)$.
A: As mentioned, there are easy counterxamples. However, it is true for UFDs since PIDs are precisely the $\rm UFDs$ of dimension $\le 1,\:$ i.e. such that prime ideals $\ne 0$ are maximal. Below is a sketch of a proof of this and closely related results.
Theorem $\rm\ \ \ TFAE\ $ for a $\rm UFD\ D$ 
$(1)\ \ $ prime ideals are maximal if nonzero
$(2)\ \ $ prime ideals are principal
$(3)\ \ $ maximal ideals are principal
$(4)\ \ \rm\ gcd(a,b) = 1\ \Rightarrow\ (a,b) = 1$ 
$(5)\ \ $ $\rm D$ is Bezout
$(6)\ \ $ $\rm D$ is a $\rm PID$  
Proof $\ $ (sketch of $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 1$)
$(1\Rightarrow 2)$ $\rm\ \ P\supset (p)\ \Rightarrow\ P = (p)$ 
$(2\Rightarrow 3)$ $\ \: $ Clear.
$(3\Rightarrow 4)$ $\ \ \rm (a,b) \subsetneq P = (p)\ $ so $\rm\ (a,b) = 1$ 
$(4\Rightarrow 5)$ $\ \ \rm c = \gcd(a,b)\ \Rightarrow\ (a,b) = c\ (a/c,b/c) = (c)$ 
$(5\Rightarrow 6)$ $\ \ \rm 0 \ne I \subset D\:$ Bezout is generated by an elt with the least number of prime factors
$(6\Rightarrow 1)$ $\ \ \rm P \supset (p),\ a \not\in (p)\ \Rightarrow\ (a,p) = (1)\ \Rightarrow\ P = (p)$
