Note that $(1,1)$ is the identity element of the ring, so any ideal containing it contains the whole ring. You're quotienting out by the subgroup generated by $(1,1)$, which is not the same as the ideal generated by it.
If $P$ is a prime ideal and $xy\in P$, then either $x\in P$ or $y\in P$, and so by induction, if $x^n\in P$, then $x\in P$. Therefore, any prime ideal containing $I$ actually contains $J=\langle 6, x-2, y \rangle$, and actually must contain one (and only one) of $J_2=\langle 2, x-2, y \rangle$ or $J_3=\langle 3, x-2, y \rangle$ (it cannot contain both, as an ideal which contains both $2$ and $3$ is trivial, and at least some definitions of prime ideal require the ideal to be proper). However, both of these ideals are maximal, as their quotients are the fields with $2$ or $3$ elements, respectively.
To see the isomorphism, we note the following fact: If $R$ is a ring, then $R[x]/(x-a)\cong R$. This follows from the first isomorphism theorem by taking the map $R[x]\to R$ sending $x$ to $a$. A slight extension of this yields that $\mathbb Z[x,y]/\langle 6, x-2, y \rangle\cong \mathbb Z[x]/\langle 6, x-2\rangle \cong \mathbb Z/\langle 6\rangle$. Actually, this gives us another perspective, as we can phrase the original problem as quotients of $\mathbb Z/\langle 6\rangle$ which are domains/fields.
For your second question, note that if an ideal contains both $y$ and $p(x,y)$ as generators, then you can replace $p(x,y)$ with $p(x,0)$ (which is the remainder upon dividing $p$ by $y$). Doing this first with $y$, then $x-2$, you have replaced $p$ with a constant polynomial.
Best Answer
That is all correct. (To get this out of the unanswered queue.)