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).
Lang uses Zorn's lemma only in the step that nonzero nonunits in a PID admit irreducible factorizations (not the uniqueness of irreducible factorizations, once we know such factorizations exist). The way he uses Zorn's lemma, I think, is excessive. What follows is how I work out the existence of irreducible factorizations when I teach the abstract algebra class.
Claim: In a PID which is not a field, any nonzero nonunit is a product of irreducibles.
We will use the following lemma (which is not Zorn's lemma).
Lemma: If $R$ is an integral domain and $a \in R$ is
a nonzero nonunit which
does not admit a factorization into irreducibles then
there is a strict inclusion of principal ideals
$(a) \subset (b)$ where $b$ is some other nonzero nonunit which
does not admit a factorization into irreducibles.
Proof of lemma: By hypothesis $a$ is not irreducible, so (since it is
neither 0 nor a unit either) there is some factorization
$a = bc$ where $b$ and $c$ are nonunits (and obviously are
not 0 either). If both $b$ and $c$ admitted
irreducible factorizations then so does $a$, so
at least one of $b$ or $c$ has no irreducible factorization.
Without loss of generality it is $b$ which has no
irreducible factorization. Since $c$ is not a unit, the inclusion
$(a) \subset (b)$ is strict. QED lemma.
Now we can prove the claim.
Proof of claim: Suppose there is an element $a$ in the PID which
is not 0 or a unit and has no irreducible factorization.
Then by the lemma there is a strict inclusion
$$
(a) \subset (a_1)
$$
where $a_1$ has no irreducible factorization.
Then using $a_1$ in the role of $a$ (and the lemma again)
there is a strict inclusion
$$
(a_1) \subset (a_2)
$$
where $a_2$ has no irreducible factorization.
This argument (repeatedly applying the lemma to
the generator of the next larger principal ideal)
leads to an infinite increasing chain of principal ideals
$$
(a) \subset (a_1) \subset (a_2) \subset (a_3) \subset \cdots
$$
where all inclusions are strict. (At this step I suppose you may say we need the Axiom of Choice to get an infinite ascending chain, but it's only countably many choices, so really not the full thrust of Zorn's lemma and in any case it feels like a less pedantic use of Zorn's lemma than the way Lang does this.) Such a chain of ideals is impossible in a PID.
Indeed, suppose a PID contains an infinite strictly increasing chain of
ideals:
$$
I_0 \subset I_1 \subset I_2 \subset I_3 \subset \cdots
$$
and set
$$
I = \bigcup_{n \geq 0} I_n.
$$
This union $I$ is an ideal. The reason is that
the $I_n$'s are strictly increasing, so any finite set
of elements from $I$ lies in a common $I_n$.
Therefore $I$ is closed under addition and arbitrary multiplications from
the ring since each $I_n$ has these properties. Because we are in a PID, $I$ is principal:
$I = (r)$ for some $r$ in the ring. But because
$I$ is the union of the $I_n$'s, $r$ is in some $I_N$.
Then $(r) \subset I_N$ since $I_N$ is an ideal, so
$$
I = (r) \subset I_N \subset I,
$$
which means
$$
I_N = I.
$$
But this is impossible because the inclusion $I_{N+1} \subset I$
becomes $I_{N+1} \subset I_N$ and we were assuming
$I_N$ was a proper subset of $I_{N+1}$.
Because of this contradiction, nonzero nonunits in a PID without
an irreducible factorization do not exist. QED
Lang's argument only uses the axiom of choice in a countable way, as above, but the way he pulls it in makes the application of Zorn's lemma feel a lot more fussy.
Best Answer
There are two parts to showing a Euclidean domain or a PID are UFDs: (i) existence of an irreducible factorization for every nonzero nonunit and (ii) essential uniqueness of the irreducible factorization (any two use the same number of irreducible factors and the irreducibles that occur in both factorizations can be matched termwise up to multiplication by a unit).
To prove (ii), the key point is that every irreducible element is a prime element, and to prove that you need to be able to write $px + ay = 1$ for any irreducible $p$ and element $a$ where $p \nmid a$; the nondivisibility implies (since $p$ is irreducible) that the only common factors of $p$ and $a$ are units, so Euclid's algorithm in a Euclidean domain lets you algorithmically solve $px + ay = 1$ for some $x$ and $y$. In a PID you'd instead observe that the ideal $(p,a)$ has to be $(1)$.
To prove (i) is a major distinction between Euclidean domains and PIDs. You can do this for Euclidean domains in a much more concrete way than for PIDs. I compare approaches for each as Theorems 4.2 and 4.3 at http://www.math.uconn.edu/~kconrad/blurbs/ringtheory/euclideanrk.pdf. You need to read Sections 2 and 3 first to see why I mean about being able to assume the "$d$-inequality" holds: a Euclidean domain does not have to have its "norm" function $d$ be totally multiplicative or satisfy $d(a) \leq d(ab)$, but you can always adjust the "norm" function to fit that inequality all the time if it doesn't at the start. (Some books make this inequality part of the definition of a Euclidean domain and some do not.) Of course in $\mathbf Z$ and $F[x]$ that inequality is true, so you save some time when proving those rings are UFDs compared to a general Euclidean domain.
The bottom line is that you definitely do not need to introduce the machinery of PIDs in order to prove rings like $\mathbf Z$ or $F[x]$ have unique factorization. After all, unique factorization in those types of rings as well as in $\mathbf Z[i]$ was known (say, to Gauss) long before there was a concept of PID. I remember being surprised when I first saw how PIDs are proved to be UFDs, since I knew the case of Euclidean domains already and the proof of part (i) for PIDs was rather more abstract than I expected.