 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 each $I_{a,b}$ is maximal.

For a maximal ideal not on this list, takeEvaluation $p(x,y)\mapsto p(i,0)$ 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 $\mathbb{R}$.