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.
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.