$\newcommand{\ideal}[1]{{\mathfrak{#1}}}$
$\newcommand{\spec}[1]{{\mathrm{Spec}}\left({#1}\right)}$
Call $\ideal{p} = (y - (x+1))$ and $\ideal{m} = (x-1,y)$, ideals of $k[x,y]$. Call further $\ideal{n} = \ideal{m} \cap A = (x_1, x_2, y)$.
Now $\ideal{p} \subseteq \ideal{m}' = (y,x+1)$ and $\ideal{m}' \cap A = \ideal{n}$. So
$$\ideal{p} \cap A \subseteq \ideal{m}' \cap A = \ideal{n} = \ideal{m} \cap A$$
Now I contend that $\ideal{p}$ is the only ideal of $k[x,y]$ that lies over $\ideal{p} \cap A$. So, as $\ideal{p} \not\subseteq \ideal{m}$ going-down is not fulfilled.
It remains to prove that $\ideal{p}$ is the only ideal of $k[x,y]$ which lies over $\ideal{p} \cap A$. This follows from the isomorphism
between $D(x_1) \subseteq \spec{A}$ and $D((x-1)(x+1)) \subseteq \spec{k[x,y]}$. This corresponds to the isomorphism
$$A_{x_1} \cong k[x,y]_{x^2-1}$$
which I have not checked in full detail, but which must exist for geometric reasons: The lines $V(x-1)$ and $V(x+1)$ in $\spec{k[x,y]}$ correspond
to the line $l$ which is the extrusion of the nodal point $(x_1=x_2 =0)$
of the nodal curve $C = V(x_2^2-x_1^2\,(x_1+1))$ along the $y$-axis. All other points get mapped 1-1. As $\ideal{p} \in D(x^2-1)$ and $\ideal{p} \cap A \in D(x_1)$, only $\ideal{p}$ can lie over $\ideal{p} \cap A$.
