= Solution
Yes. Let $\mathfrak m$ be maximal and set
$$
L=\mathbb C[T_1,T_2,\ldots]/\mathfrak m.
$$
Then $L$ is a field generated as a $\mathbb C$-algebra by countably many elements. Since the polynomial ring in countably many variables has a countable monomial basis, $L$ has at most countable dimension as a $\mathbb C$-vector space.
Suppose $t\in L$ were transcendental over $\mathbb C$. The family
$$
\left\{\frac1{t-x}:x\in\mathbb C\right\}
$$
would be linearly independent over $\mathbb C$. Indeed, after multiplying a finite relation by $\prod_j(t-x_j)$, evaluation at $t=x_i$ forces the $i$th coefficient to vanish. This would be an uncountable linearly independent subset of the countable-dimensional vector space $L$, a contradiction.
Thus $L/\mathbb C$ is algebraic. Since $\mathbb C$ is an <algebraically closed field>, $L=\mathbb C$. If $x_i$ is the image of $T_i$, the quotient map is evaluation at $(x_1,x_2,\ldots)$ and
$$
\boxed{\mathfrak m=(T_1-x_1,T_2-x_2,\ldots)}.
$$
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