Below is a $ $ complete $ $ proof from first principles - easily comprehensible to a high-school student.
We show $\rm\:(2,x)\ =\ (f)\ $ in $\rm\:\mathbb Z[x]\: $ yields a parity contradiction, by simply evaluating polynomials.
$\rm\ \ f\ \in\ (2,x)\, \Rightarrow\, f\ =\ 2\, G + x\, H.\: $ Eval at $\rm\: x = 0\ \Rightarrow\ \color{#0a0}{f(0)} = 2\,G(0) = \color{#c00}{2n}\:$ for some $\rm\: n\in \mathbb Z$
$\rm\ \ 2\ \in\ (f)\ \Rightarrow\ 2\ =\ f\, g\:\ \Rightarrow\ deg(f)\ =\ 0\:\ \Rightarrow\:\ \color{#c00}f\ =\ \color{#0a0}{f(0)}\ =\ \color{#c00}{2n}$
$\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 $\, \Rightarrow\!\Leftarrow$
Remark $\ $ The above proof works over any domain where $\,2\ne 0\,$ and $\rm\,2\,$ is not a unit. $ $ i.e. $\rm\:2\nmid 1.\:$ In particular, it works over any domain with a nontrivial sense of parity, i.e. having $\rm\:\mathbb Z/2\:$ as ring image, e.g. the Gaussian integers, or the rationals writable with odd denominator - see this post. Conversely, the result is false if $\rm\:2 = 0\,$ or a unit since then $\rm\,(2,x) = (x)\,$ or $\,(1)\,$ is principal.
Further, the proof still works if we replace $\,2\,$ by any element $\,c\,$ of the coefficient domain $\,D,\,$ yielding: $\ (c,x)\,$ is principal in $\,D[x]\iff c=0\,$ or $\,c\,$ is a unit.