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: 
One possible definition of $\mathbf C$ is that it is the splitting field of the polynomial $X^2 +1$. As such, it is isomorphic to the quotient ring $\mathbf R[X]/(X^2+1)$ (polynomials with real coefficients, modulo the ideal generated by $X^2+1$). The imaginary number i is then simply the congruence class of X.
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:
$$\mathbb{C}\cong\mathbb{R}[x]/(x^2+1)$$