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: 
The $\mathbb R[x] / (x^2 + 1)$ solution is what leapt to everyone's mind, but there is an even simpler solution:

$\Bbb C[x]/(x)\cong \Bbb C$.

In both cases, the polynomial whose ideal is being modded out is a maximal ideal of the ring (it would have to be maximal, after all), so it is very much like the integers modulo a prime, as you wished.

Or, for that matter, you could even just say $\Bbb C/\{0\}\cong \Bbb C$, if you don't mind a trivial solution.
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:
$$\mathbb{C}\cong\mathbb{R}[x]/(x^2+1)$$