Questions about following proof regarding why $\mathbb{Z}[x]$ is not a principal ideal domain 
Prove that $\Bbb{Z[x]}$ is not a principal ideal domain.


Proof: Consider the ideal $I=\langle x,2\rangle$. We’ll show that this ideal is not principal. First note that I is not equal to $\Bbb{Z[x]}$ because 1 is not in I. If it were then $1=xf(x)+2g(x)$ hx, 2i because if it were then $1 = xf(x) + 2g(x)$ for f(x), g(x) ∈ Z[x], but xf(x) + 2g(x) has even constant term.
Then suppose $I=\langle p(x) \rangle$  for some p(x) ∈ Z[x]. Then we must have x = p(x)f(x) and 2 = p(x)g(x) for some f(x), g(x) ∈ Z[x]. But the second implies that p(x) must be a
constant polynomial, specifically p(x) = −2, −1, 1 or 2. We can’t have p(x) = ±1 because then
I = Z[x] so p(x) = ±2. But then x = ±2f(x), a contradiction since ±2f(x) has even coefficients.

I don't understand this part But then x = ±2f(x), a contradiction since ±2f(x) has even coefficients. What is the thing about the even coefficients, and why xf(x) + 2g(x) has even constant term in the first part? How do they lead us to the contradiction?
 A: If we assume that $I = \left<p(x)\right>$ for some $p(x) \in \Bbb{Z}[x]$, that means that we can write $\textit{any}$ element of $I$ as a product of $p(x)$ and another element of $\Bbb{Z}[x]$. In particular, we then could write the two polynomials $x$ and $2$ in this way, i.e.,
$$x = f(x)p(x)$$
$$2 = g(x)p(x)$$
for some $f(x),g(x) \in \Bbb{Z}[x]$. Let us first analyze the second equality. $2 = g(x)p(x)$ implies that $p(x)$ must be a constant polynomial, since otherwise the degree would not match. Further, we can also see that (depending on what $g(x)$ would be) $p(x)$ could in any case only take the values $\pm1, \pm2$.
We already know that it cant be $\pm1$ since then we had $I = \Bbb{Z}[x]$, which you have already excluded. So let us now check the first of the above equations when we assume that $p(x) = \pm2$. If that is the case, then $f(x)p(x)$ for sure has even coefficients, no matter what $f(x)$ is. But that's a problem, since we have that $x = f(x)p(x)$, and $x$ doesn't have even coefficients! Here is where the contradiction arises.
A: The proof shows that $x=\pm 2f(x)$. But no matter what $f(x)$ is, after multiplying by $\pm 2$, all of the coefficients of $f(x)$ will be even (as a multiple of $2$). But the coefficient of $x$ is $1$, which is odd. So if $x=\pm 2f(x)$, then $1$ is even, a contradiction. Does this make sense?
