It is a maximal ideal because $x^2+x+1$, a quadratic polynomial with no root in $\mathbf Z_2$, is irreducible in $\mathbf Z_2[x]$ and in PIDs irreducible elements generate maximal ideals.
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
If $\left<x^2+y^2+z^2\right>$ is not a prime ideal, there should exist $g,h\in \mathbb{R}[x,y,z]$ s.t. $x^2+y^2+z^2\mid gh$, $x^2+y^2+z^2\nmid g,h$.
Since $\mathbb{R}[x,y,z]$ is a UFD, this implies there exist nonunit $\alpha,\beta\in\mathbb{R}[x,y,z]$ s.t. $\alpha\beta=x^2+y^2+z^2$.
$\alpha,\beta\neq 0$, so $\alpha,\beta$ not being unit implies $\deg(\alpha),\deg(\beta)>0$. Since $\deg(\alpha\beta)=2$ and $\mathbb{R}$ is integral, $\deg(\alpha)+\deg(\beta)=2\Rightarrow \deg(\alpha)=\deg(\beta)=1$.
We can write $\alpha=ax+by+cz$.
Now, observe that for every $(x,y,z)\neq(0,0,0)$, $x^2+y^2+z^2\neq 0$. Since we can find nontrivial $(x,y,z)$ s.t. $ax+by+cz=0$, this is a contradiction.