Prove that $\mathbb{Z}[i]/\langle 1+i \rangle \cong \mathbb{Z}/2\mathbb{Z}$

You are almost there. Elements in $\langle 1+i \rangle$ are exactly those that you described: those of the form $(1+i)(c+di)$. These are clearly in the kernel (because $1+i$ is in the kernel and you already showed $\phi$ takes products to products).

The other containment is more interesting. Given $a + bi$ in the kernel, you need to show that you can divide it by $1+i$. More precisely, you need to show that there exists $c + di \in \mathbb{Z}[i]$ such that $a + bi = (1+i)(c+di)$.

My hint would be that you can guess the formula for $c+di$ by simplifying the fraction $\frac{a+bi}{1+i}$. Be sure to verify that everything that shows up in your simplified fraction is an integer. To do this you will use that $a+bi$ is in the kernel, i.e. that $a+b$ is even.

The construction of the homomorphism is good. Since $\phi(1+i)=[0]_2$, you know that $\langle 1+i\rangle\subseteq\ker\phi$. Conversely, if $a+bi\in\ker\phi$, you have $a+b=2c$ for some $c$ and $$ a+bi=(1+i)(x+yi) $$ where $x+y=b$, $x-y=a$ (and the solution exists) because $a+b$ and $a-b$ are even.

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However, you're not using the homomorphism theorem. For this you can proceed in several steps. Consider the unique homomorphism $\alpha\colon\mathbb{Z}[X]\to\mathbb{Z}/2\mathbb{Z}$ such that $\alpha(X)=[1]_2$ ($X$ is an indeterminate). The ideal $\langle X^2+1\rangle$ is contained in the kernel of $\alpha$, since $\alpha(X^2+1)=[1]_2^2+[1]_2=[0]_2$.

The homomorphism theorem provides a surjective homomorphism $\beta\colon\mathbb{Z}[X]/\langle X^2+1\rangle\to\mathbb{Z}/2\mathbb{Z}$, that is, $\beta\colon\mathbb{Z}[i]\to\mathbb{Z}/2\mathbb{Z}$.

You can notice that this homomorphism is exactly your $\phi$. Then you can proceed as before.

If you already know that $\langle 1+i\rangle$ is maximal, the proof ends with the observation that $\langle 1+i\rangle\subseteq\ker\beta$.