Question: Can the complex numbers be realised as some $R/M$ where $R$ a ring and $M$ a maximal ideal like the integers modulo some prime? I understand that unlike the latter case, such a maximal ideal would need to partition the ring into infinitely many cosets.

Answer: 
Yes, $\mathbb{C} = \mathbb{R}[x] / \langle x^2 + 1 \rangle$ is the standard contruction, where $\mathbb{R}[x]$ is the ring of polynomials in one variable with real coefficients; and $\langle x^2 + 1 \rangle$ is the ideal generated by the polynomial $x^2 + 1$. 

If you think about it, since $x^2 + 1$ generates the ideal, it's true that

$$x^2 + 1 \equiv 0 \pmod{x^2 + 1}$$

and therefore $x^2 \equiv-1$ so that the equivalence class of the polynomial $x$ is a square root of $-1$. 

By mapping $1$ to $1$ and $i$ to $x$ you can get an isomorphism. 

[Markup question, Why do I have an extra space in my mod expression above?]
I would like to add that the standard construction $\mathbb{R}[x]/(x^2+1)$ mentioned by several of the answers here can be thought of as a private case of a more general theorem regarding field extensions:

**Theorem**: If $E/F$ is a field extension, and some element $a\in E$ is algebraic over $F$, then $F(a)$ (the field generated by $a$ over $F$) is isomorphic to $F[x]/(m\_a)$ (where $m\_a$ is the minimal polynomial of $a$).

**Proof Idea**: We can define a homomorphism from $F[x]$ to $F(a)$ given by $f(x)\mapsto f(a)$, and then use the first isomorphism theorem for rings.

In our case, it is obvious that $\mathbb{C}=\mathbb{R}(i)$, and $m\_i=x^2+1$, and we immediately get:
$$\mathbb{C}\cong\mathbb{R}[x]/(x^2+1)$$