Dear Arturo,
The exercise in question is actually a theorem of Kaplansky. It appears as Theorem 5 on page 4 of his Commutative Rings. [I was not able to tell easily whether the result appears for the first time in this book.] The proof is reproduced in Section 10 of an expository article I have written [but probably not yet finished] on factorization in integral domains:
http://alpha.math.uga.edu/~pete/factorization.pdf
Regarding your second question, there has been some work on understanding Euclidean domains from more intrinsic perspectives. Two fundamental articles are:
Motzkin, Th. The Euclidean algorithm. Bull. Amer. Math. Soc. 55, (1949). 1142--1146.
http://alpha.math.uga.edu/~pete/Motzkin49.pdf
Samuel, Pierre About Euclidean rings. J. Algebra 19 1971 282--301.
http://alpha.math.uga.edu/~pete/Samuel-Euclidean.pdf
I have not had the chance to digest these papers, so I'm not sure if they directly answer your question (maybe not, but I think they will be helpful).
Even in the noncommutative case, we can show that a local ring whose maximal ideal $\mathfrak{m}$ is left principal is a left PIR if and only if $\bigcap_{i \geq 0} \mathfrak{m}^i = 0$ (the conclusion of Krull's intersection theorem).
Recall that a noncommutative ring $R$ is local if the non-units form a left ideal $\mathfrak{m}$ (and in this case $\mathfrak{m}$ is in fact a two-sided ideal). Suppose this maximal ideal is left-principal, say $\mathfrak{m} = Rt$.
The "if" direction follows the same proof as in the commutative case.
For the "only if" direction, suppose that $R$ is a left PIR. Let $I = \bigcap_{i \geq 0} \mathfrak{m}^i$. Since $R$ is a PIR, we have $I = Rx$ for some $x$. Similar to the proof of Krull's intersection theorem in the commutative case, show that $tI =I$ (if $z \in I$, then $z = tz'$ for some $z'$; if $z' \notin I$, then $z' \notin \mathfrak{m}^j$ for some $j$, but then $z \notin \mathfrak{m}^{j+1}$, contradicting $z \in I$). Then there is some $y \in I$ such that $x = ty$. Since $y \in I=Rx$, we have $y=rx$ for some $r$, and thus $x = trx$, or $(1-tr)x=0$. Since $tr \in \mathfrak{m}$, $1-tr$ is a unit, and thus $x=0$ and $I = Rx = 0$.
(I don't even think you need Noetherian to prove this; in case $R$ ends up being a left PIR, Noetherian seems to follow.)
Best Answer
I assume that by a valuation domain you mean an integral domain $R$ with fraction field $K$ such that: for all $x \in K^{\times}$, at least one of $x,x^{-1}$ lies in $R$.
In this case, I believe the answer is no. Let $R$ be any valuation domain whose value group $K^{\times}/R^{\times}$ is isomorphic, as a totally ordered abelian group, to $\mathbb{Z} \times \mathbb{Z}$ with the lexicographic ordering. (It is known that every totally ordered abelian group is the value group of some valuation domain, e.g. by a certain generalized formal power series construction due to Neumann.) In this case, the maximal ideal consists of all elements whose valuation is strictly greater than $(0,0)$, but the valuation of any such element is at least $(0,1)$ and therefore any element of valuation $(0,1)$ gives a generator of the maximal ideal.
For some information on valuation rings, see e.g. Section 17 of
http://alpha.math.uga.edu/~pete/integral.pdf