= Solution
Evaluation at a real point defines a surjective <ring homomorphism> $p\mapsto p(a,b)$ from $\mathbb R[x,y]$ to $\mathbb R$. Its <kernel> is $(x-a,y-b)$: subtracting the constant $p(a,b)$ expresses the remainder as a multiple of $x-a$ plus a multiple of $y-b$, by successive <polynomial division>. The <first isomorphism theorem> therefore identifies the <quotient ring> with the <field> $\mathbb R$, proving \b[each $I_{a,b}$ is maximal].
For a <maximal ideal> not on this list, take
$$
\boxed{J=(x^2+1,y).}
$$
Evaluation $p(x,y)\mapsto p(i,0)$ is onto $\mathbb C$, and its <kernel> is $J$: first subtract a multiple of $y$, then divide the remaining polynomial in $x$ by $x^2+1$. Hence $\mathbb R[x,y]/J\cong\mathbb C$, a <field>. Finally, if $J=I_{a,b}$ for real $a,b$, the member $x^2+1\in J$ would vanish at $(a,b)$, requiring $a^2+1=0$, which is impossible over $\mathbb R$.
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