When a prime ideal in polynomial ring over integers is principal 
While dealing with a question about a prime ideal $I\subset\mathbb{Z}[x]$ (with $0$ as the only constant polynomial in $I$) I was asked to show that there exists $f(x)\in\mathbb{Z}[x]$ such that $I=\left<f(x)\right>$ and got stuck.

How do I know that some ideal is generated by only one element?
What have i tried:
i) Looked at $I$ and tried to adapt the proof that integers is a principal ideal domain and and so as $I$ is an ideal of itself I'd be done, but no success.
ii) Ignored one operation and tried to see I as a subgroup, but that was a rather unsuccessful strategy.
I appreciate any hints and book references where i can find proper explanations.

let $P$ and $Q$ are in $I$ of minimum degree so $degP=degQ$. in
the case one is unitary  for example $P$, then $I=(P)$ this by
Euclidean division and end. so we suppose that all $P$,for which
degree are the  minimum  degree of element in $I$,  are not
unitary.
so we suppose  $P$ and $Q$ are not unitary but primitive because
we can always  write $P=gcd(a_i,i=0,...n)P_1$ with $P_1$ primitive
(hypothesis that $I$ prime and no constant polynomial  other zero
in $I$).
so we prove that $P=Q$ and $I=(P)$ .
the polynomial $b_nP-a_nQ$ is
in $I$ where $a_i$ and $b_i$ are the coefficient of $P$ and $Q$
respectively. so this polynomial must be zero (by minimal degree
of element of $I$) and so $ P=Q$ this prove that each element  $Q$
in $I$ s.t   $degre Q= deg P$ that $ Q=cP$ for some constant c.
Now we prove by recurrence on degree of  $Q$ that $Q= C(X)P(X)$.
so suppose $degQ=(degP)+1=n+1$.
$Q=b_{n+1}X^{n+1}+...+b_0\in I$ so $a_nQ-b_{n+1}XP\in I $ of degre
n (note: if equal to zero then $a_nb_0=0$ and $Q=XL$ with
$degrL=n$ so $L=cP$, because $X$ unitary so $X$ is not in $I$ ) so
is of form $cP$, by substitution $X=0$, we conclude that
$a_nb_0=ca_0$ and this applique $b_0=a_0b'_0$ ($a_0, a_n$ are
coprime numbers ). so $Q-b'_0P=c'XP$ and this applique $Q=AP$.
suppose that each $R\in I$ of degree $i$,  $n< i< m$ is in the
forme $AP$. (also we use the same technique as in the case $i=n+1$
we obtain the result).
let $Q\in I$ s.t $degrQ=m$, then
$a_nQ-b_mXP\in I $ of degree $m-1$ (suppose not zero as in the
above case) and so is of the  forme $C(X)P(X)$; by tacking $X=0$,
we have $a_nb_0=c_0a_0$ so $b_0=a_0b'_0$ and
$Q(X)-b_0P(X)=XL(X)\in I$, so $L=bP$ and $Q=AP$.
