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?

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?
