Prove $I$ is non-principal ideal of $\mathbb{Z}[x]$? I'm new to algebra and got stuck with concept of ideals.

The question is to prove that

$$I = \left\{ {{a_0} + {a_1}x +  \cdots  + {a_n}{x^n} \mid {a_i} \in \mathbb{Z},{a_0} \in 2\mathbb{Z}} \right\}
$$
is a non-principal ideal of $\mathbb{Z}[x]$.
I tried to show for any $f \in \mathbb{Z}\left[ x \right]$ the principal ideal $fR \ne I$.
So I considered three different cases:
1) $f$ is an odd constant polynomial.
2) $f$ is an even constant polynomial.
3) $f$ is a non-constant polynomial.
Then I had,
$$\begin{array}{l}
aR = a\left( {{\alpha _0} + {\alpha _1}x +  \cdots {\alpha _n}{x^n}} \right)\\
bR = b\left( {{\alpha _0} + {\alpha _1}x +  \cdots {\alpha _n}{x^n}} \right)\\
\left( {x + 1} \right)R = {\alpha _0} + \left( {{\alpha _0} + {\alpha _1}} \right)x + \left( {{\alpha _1} + {\alpha _2}} \right){x^2} +  \cdots  + \left( {{\alpha _{n - 1}} + {\alpha _n}} \right){x^n}
\end{array}
$$
where ${\alpha _i} \in \mathbb{Z}$ and $a$ is an even constant and $b$ is an odd constant. 
But I cannot distinguish any difference between $fR$ and $I$ with above expressions.
Or is it just enough to say that
$$\begin{array}{l}
2 \cdot R = 2{\alpha _0} + 2{\alpha _1}x +  \cdots  + 2{\alpha _n}{x^n}\\
\therefore 2{\alpha _1},2{\alpha _2}, \cdots ,2{\alpha _n} \ne \mathbb{Z}\\
\\
1 \cdot R = {\alpha _0} + {\alpha _1}x +  \cdots  + 2{\alpha _n}{x^n}\\
\therefore {\alpha _0} \ne 2\mathbb{Z}\\
\\
\left( {1 + x} \right) \cdot R = {\alpha _0} + \left( {{\alpha _0} + {\alpha _1}} \right)x +  \cdots  + \left( {{\alpha _{n - 1}} + {\alpha _n}} \right){x^n} + {\alpha _n}{x^{n+1}}\\
\therefore {x^{n+1}} \notin I
\end{array}
$$
 A: You can note that $2\in I$. So, if $I$ is principal, say $I=(f)$, then $2$ should be divisible by $f$. But there are not too many polynomials that divide $2$ (namely, you have $1$ and $2$), and none of the ideals $(1),(2)$ equals $I$.
A: Notice $\rm\,I = (2,x).\,$ We show $\rm\:I\ =\ (f)\ $ in $\rm\:\mathbb Z[x]\: $ yields a parity contradiction.
$\rm\ \ f\ \in\ (2,x)\ \Rightarrow\ f\ =\ 2\, G + x\, H\:.\: $ Eval at $\rm\: x = 0\ \Rightarrow\ \color{#c00}{f(0)}\ =\ 2\ G(0)\ =\ \color{#c00}{2n}\:$ for  $\rm\: n\in \mathbb Z$
$\rm\  \ 2\ \in\ (f)\ \Rightarrow\ 2\ =\ f\, g\:\ \Rightarrow\,\ deg\ f\ =\ 0\:\ \Rightarrow\:\  \color{#c00}{f\ =\ f(0)\ =\ 2n}$
$\rm\ \ x\ \in\ (f)\ \Rightarrow\ x\ =\ \color{#c00}f\, h\ =\ \color{#c00}{2n}h.\,\ $ Eval at $\rm\ x = 1\ \Rightarrow\ 1\: =\ 2n\,h(1)\ \Rightarrow\ 1\:$ is even $\, \Rightarrow\Leftarrow$
A: I propose you the following hinted way: 
Suppose $\;I=\langle\; f(x)\;\rangle\;,\;\;f(x)\in\Bbb Z[x]\;$ . Since $\;\deg(kf)\ge\deg f\;\;\forall \;k(x)\in\Bbb Z[x]$ , we get that all the elements in $\;I\;$ are either the zero polynomial or else polynomials of degree at least $\;\deg f\;$ .
Since $\;2\in I\;$ the above means $\;\deg f=0\;$ and thus $\;f(x)=k\;,\;\;k\in\Bbb Z\;$ is a constant polynomial . But we also have that $\;x+2\in I\;$ , so it must be that $\;x+2=kg(x)\;$ , for some $\;g(x)\in\Bbb Z[x]\;$ .
End the argument now.
