Please check my proof of the following:
If $V$ is a finite dimensional $\mathbb{C}[[x]]$-module (over $\mathbb{C}$), then the action of $x$ is nilpotent.
Let $X \in M_{n \times n} (\mathbb{C})$ be the matrix corresponding to $x$. Suppose $\lambda \neq 0$ is an eigenvalue for $X$ with eigenvector $v \neq 0$. (In a finite dimensional vector space, $X$ must have an eigenvalue.) Then $$\left(\frac{1}{\lambda} X - I_n\right)(v)=0.$$
However, the element $\frac{1}{\lambda} x - 1 \in \mathbb{C}[[x]]$ is invertible, in particular, $$-\left(1+\frac{1}{\lambda}x + \frac{1}{\lambda^2}x^2 + \dots \right)\left(\frac{1}{\lambda}x-1\right) = 1.$$ But now let both sides act on $v$, and we see that $$0=v,$$ a contradiction. Thus, $X$ has no nonzero eigenvalue, which (in the finite dimensional case!) implies that $X$ is nilpotent.
This result is quoted in Mazorchuk's "Lectures on $\mathfrak{sl}_2(\mathbb{C})$-Modules" in Section 3.7 in the form:
In order for finite dimensional $\mathbb{C}[[x]]$ modules to be well-defined, $x$ must be nilpotent.
Just wanted to make sure I was understanding this correctly. Also, if $x$ is nilpotent, why talk about power series at all? Isn't a $\mathbb{C}[[x]]$-module just the same as a $\mathbb{C}[x]/ (x^n)$ module for some $n$? Perhaps, since he's talking about the category of all $\mathbb{C}[[x]]$ modules, this is easier than saying the class of $\mathbb{C}[x]/ (x^n)$ modules for every $n$?
As you noted, for every nonzero $\lambda$ you have that $1-\lambda x$ is invertible in your ring, so it follows that $x$ has no non-zero eigenvalue on $V$, since the image of this element has then non-zero determinant. Since $V$ is finite dimensional, this means of course $x$ is nilpotent. This also proves, as you noted, that any map $\mathbb C[[x]]\to \operatorname{End}(V)$ where $V$ is finite dimensional factors through one of the projections $\mathbb C[[x]]\to \mathbb C[x]/(x^n)$.