A variation on $6=2 \cdot 3=(1+\sqrt {-5})(1-\sqrt {-5})$ In the integral domain $\mathbb{Z}[\sqrt {-5}]$, we have $6=2 \cdot 3=(1+\sqrt {-5})(1-\sqrt {-5})$, so it is not a UFD (two different decompositions of $6$).
$2$ divides $6$, but $2$ does not divide $1+\sqrt {-5}$ nor $1-\sqrt {-5}$.

(1) Could one present an example of an integral domain $D$ in which an element $d \in D$ divides $a^n$ but $d$ does not divide $a$? where $a \in D$ and $n \in \mathbb{N}$.

Define property $P$ in an integral domain the following property: 
If an element $d \in D$ divides $a^n$, then $d$ divides $a$, where $a \in D$ and $n \in \mathbb{N}$.
(Is it true that Property $P$ characterizes GCD domains?).

(2) What is known about integral domains satisfying property $P$?
  Is it true that $\mathbb{Z} [\sqrt {-5}]$ satisfies property $P$?

Thank you very much!
 A: Maybe I'll think of an example for $\mathbb{Z}[\sqrt{-5}]$ after I log off. But I thought of one in $D = \mathcal{O}_{\mathbb{Q}(\sqrt{-15})}$ before I even finished reading your question.

(1) Could one present an example of an integral domain $D$ in which an element $d \in D$ divides $a^n$ but $d$ does not divide $a$? where $a \in D$ and $n \in \mathbb{N}$.

Consider $$d = \frac{1}{2} + \frac{\sqrt{-15}}{2}$$ (which is an algebraic integer with minimal polynomial $x^2 - x + 4$), $a = 2$, $n = 4$. Then $d \mid a^n$, but $d \nmid a$, since $$\frac{2}{\frac{1}{2} + \frac{\sqrt{-15}}{2}} = \frac{1}{4} - \frac{\sqrt{-15}}{4}$$ is an algebraic number but not an algebraic integer (minimal polynomial $2x^2 - x + 2$). You'll see much the same thing with the conjugate of $d$.
Actually, not sure I'll be able to squeak it in before I have to go, but I just remembered about $(2 - \sqrt{-5})(2 + \sqrt{-5}) = 3^2$, which the
As for property $P$, I'm inclined to say that it is true of GCD domains but not others, though I could be overlooking something commutative that I ought to know about but can't remember at the moment.
In either case, I can tell you that $\mathbb{Z}[\sqrt{-5}]$ doesn't have property $P$, since then of
