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?]
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.