If $R=\mathbb{R} [x_1 , x_2, \ldots ]$ and $I=(x_1^2, x_2^2, x_3^2, \ldots )$ then does $R/I$ contain only one maximal ideal? Just for the sake of clarity, $R$ is the ring of polynomial of infinitely many variables over $\mathbb{R}$.
I claim that $R/I$ does not contain only one maximal ideal. If the opposite was true then $R/I$ would be a field and hence $I$ would be a maximal ideal of $R$. But, $I \subset (x_1 , x_2 , x_3, \ldots ) \subset R$ and hence $I$ cannot be a maximal ideal!
But it seems my claim isn't correct. Can someone show me where I am going wrong?
$R/I$ having only one maximal ideal does not mean it is a field. $R/I$ having only one proper ideal would mean it is a field. For example, $R=\mathbb Z$, $M=2\mathbb Z$ and $I=4\mathbb Z$, we have $R/I$ has exactly one maximal ideal, but $R/I$ is not a field.
In fact, original proposition is true.
Firstly as you noticed, $M=(x_1, x_2,\ldots)$ is an important ideal, and indeed it is maximal since the quotient by this ideal is isomorphic to $\mathbb R$.
Secondly if $M'$ is any maximal ideal, it is also necessarily prime. Since $M'$ contains $x_i^2$ for every $i$ and it is also prime, it must also contain $x_i$ for every $I$. That means $M'\supseteq M$. But by maximality of $M$, $M=M'$. So there is only one maximal ideal.