Units/zero divisors in $R[x]$ 
Let $R$ be the set of integers modulo $9$, $\mathbb{Z}_9$, and define $R[x]$ to be the corresponding polynomial ring. Determine whether $x, 3x, 1 + x$, and $1 + 3x$ are units or zero divisors.

My attempt:

$3x$ is a zero divisor since $3x \cdot 3 = 9x = 0$ in mod $9$. $1 + 3x$ is a unit since
\begin{align*}
(1 + 3x)(1 - 3x) = 1 - 3x + 3x + 9x^2 \equiv 1.  
\end{align*}

I cannot figure out $x$ and $1 + x$. By trial and error, I believe they are neither units nor zero divisors, but I cannot figure out how to prove it.
REVISED ATTEMPT:

Consider a non-zero polynomial $p(x) = a_0 + a_1 x + \ldots + a_n x^n$ where $a_n \neq 0$ for some $n \geq 0$. Then
\begin{align*}
x p(x) & = x (a_0 + a_1 x + \ldots + a_n x_n) \\
& = a_0 x + a_1 x^2 + \ldots + a_n x^{n+1}.
\end{align*}
Since $a_n \neq 0$, $p(x)$ is a polynomial of degree $n + 1$. Since $n \geq 0$, this implies that $p(x)$ is a polynomial of degree $\geq 1$. Therefore, it cannot equal either $0$ or $1$, so it is neither a zero divisor or a unit.


What about $1+x$? Let's try multiplying by a nonzero polynomial
$$
(1+x)(a_0+a_1x+\dots+a_nx^n)=
a_0+(a_1+a_0)x+(a_2+a_1)x^2+\dots+(a_n+a_{n-1})x^n+a_nx^{n+1}
$$
If $1+x$ is a unit, we should be able to find
$$
a_n=0,\quad a_n+a_{n-1}=0,\quad\dots,\quad a_1+a_0=0,\quad a_0=1
$$
and you immediately see the contradiction.
Can $1+x$ be a zero divisor? Again not, because we'd need $a_0=0$, so also $a_1=0$ and so on up to $a_n$.
What's the difference between $1+x$ (neither a unit nor a zero divisor) and $1+3x$ (a unit)? That $3x$ is nilpotent.
Your attempt for $x$ is good: if $p(x)$ has degree $n\ge0$, then $xp(x)$ has degree $n+1$.
Any time a polynomial $q(x)$ has a unit as its leading term, the degree formula holds $\deg(q(x)p(x))=\deg q(x)+\deg q(x)$. Prove it. This provides a different proof of the fact that $1+x$ is neither a unit nor a zero divisor.
