Finding all ideals in $\mathbb{C}[[x]]$

I am currently trying to find all ideals in $\mathbb{C}[[x]] = \{\sum_{i=0}^\infty a_ix^i : a_i \in \mathbb{C}\}$, that is, the ring of Taylor series with complex coefficients. I know that since the set of complex numbers $\mathbb{C}$ forms an integral domain under addition and multiplication, so does $\mathbb{C}[[x]]$. I also proved that $R[[x]]^\times = \{\sum_{i=0}^\infty a_ix^i : a_i \in R, a_0 \in R^\times\}$, where $R$ is any ring. I wonder if any link can be made between the units of $R[[x]]$ and its ideals as they are part of the same assignment question.

I was thinking of using the First Isomorphism Theorem for Rings since the kernel of a ring homorphism $\phi : R \to S$ is an ideal of $R$, although I am note quite sure this will give me all ideals. My question is: In general, what is the best approach to find ideals in an infinite ring? Finite rings are usually much easier, but it is the first time I encounter this question for the infinite case.

Thanks a lot for your help.

First of all, $(x)$ is a maximal ideal, as the quotient is isomorphic to $\mathbb C$. Your description of the units can be rephrased as $\mathbb C[[x]]^\times=\mathbb C[[x]]\setminus (x)$. This says that $(x)$ is the unique maximal ideal. So the ring is local. Moreover, it has dimension one, as the unique chain of prime ideals that you can imagine in this ring is $$(0)\subset (x)\subset \mathbb C[[x]],$$ which has lenght one. Finally, the ring is regular as a local ring, because $$\dim_\mathbb C(x)/(x)^2=\dim \mathbb C[[x]].$$ So we have a regular local ring of dimension one, in other words a discrete valuation ring. Its ideals are exactly the powers of the maximal ideal.
In general, ideals of a ring $R$ are exactly the kernels of ring homomorphisms with source $R$.
I do not think there is a standard strategy to list all the ideals in a ring $R$. But you could start by listing all maximal ideals, and then all prime ideals, like we did: in this case the only primes were $(0)$ and $(x)$, and $(x)$ was maximal. Then you may ask whether $R$ is a domain (i.e. $(0)$ is prime), a principal ideal domain, a Dedekind ring, what is the dimension... all these information give some info about ideals. But not so much information: if the ring is just a bit nasty, this problem is hard. Ideals in a ring form a lattice, and very easily this is a complicated object (sorry for the wordplay).
Since you know how are the units, in the same way you can show that if an ideal $I$ contains a power series whose lowest nonzero coefficient is $a_{i}$, then it contains $x^{i}$ (it is the same proof, you just write the power series as $x^{i}\sum_{j=0}^{\infty}a_{i+j}x^{j}$). So you can check easily that the ideals are of the form $\left(x^{i}\right)$ for some $i$. In this way you have also that the ring is local.