As noted in a comment, we need to add the hypothesis that $\operatorname{ann}(x)=\{x,-x,0\}$ and that these elements are distinct. Notice that we have $x^2=0$.
Let $r\in R$. Then $rx$ annihilates $x$, so $rx$ must be one of $x,-x,0$. If $rx=x$, then $(r-1)x=0$, and so $r-1$ is one of $x,-x,0$. Thus $r$ is one of $1+x,1-x,1$. After we consider the cases $rx=-x$ and $rx=0$, we see $R=\{-1,0,1,-1+x,x,1+x,-1-x,-x,1-x\}$. This shows that $R$ is generated as a ring by $x$.
There are now two ways to finish.
(Way 1) With a little more work, we can see that these elements are all distinct. [It helps to notice $2x\ne 0$ since $x\ne -x$, so $\pm2$ is not in $\operatorname{ann}(x)$.] The commutative rings with $9$ elements are $\mathbb{F}_9,\mathbb{Z}_3\times\mathbb{Z}_3,\mathbb{Z}_3[x]/(x^2),\mathbb{Z}_9$, and the first two have no nilpotent elements.
(Way 2) $2x\in\operatorname{ann}(x)$, so $2x$ is one of $0,x,-x$. If $2x=0$, then $x=-x$ and if $2x=x$, then $x=0$. Thus we must have $2x=-x$, i.e., $3x=0$. This in turn implies $3$ (as in $31_R$) is one of $0,x,-x$.
If $3=0$, then $R$ is an algebra over $\mathbb{Z}_3$, generated by $x$ with $x^2=0$. This is precisely the algebra $\mathbb{Z}_3[x]/(x^2)$.
If $x=\pm3\ne0$, then we have $0=3x=\pm 9$, so $R$ is an algebra over $\mathbb{Z}_9$ generated by $\pm3$. This is precisely $\mathbb{Z}_9$.
