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?]
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.