Consider a quiver $Q$ that is not of finite type. Can you construct a free $k[t]$-representation $M$ of $Q$ such that the $k$-representations $M\otimes_{k[t]}k[t]/(t-\lambda)$ are indecomposable and belong to infinitely many isomorphism classes (you may assume $k$ algebraically closed, and you are free to choose a dimension vector)?
This should follow from the tame-wild dichotomy, but I would like to see a more direct proof.
You can do this more or less "by hand" if you're willing to use Gabriel's theorem which classifies quivers of finite type. I will assume that $Q$ is connected (to simplify) and that $k$ is algebraically closed (it could just be infinite for this proof).
The first step is an application of Gabriel's theorem: if $Q$ is of finite type, then it is an orientation of a Dynkin diagram of type $A,D$ or $E$.
The second step is combinatorial: if $Q$ is not an orientation of a Dynkin diagram of type $A,D$ or $E$, then it contains as a subquiver an orientation of a diagram of type $\tilde{A}$, $\tilde{D}$ or $\tilde{E}$. This is proved by a straightforward case by case analysis.
The third step is the observation that if $Q'$ is a subquiver of $Q$, then any indecomposable representation of $Q'$ induces an indecomposable representation of $Q$, and this induction procedure preserves isomorphism classes. Therefore, to prove the statement of the question, it suffices to prove it for quivers which are orientations of diagrams of type $\tilde{A}$, $\tilde{D}$ or $\tilde{E}$.
The last step is then devoted to finding a representation $M$ as in the statement of the question for quivers of type $\tilde{A}$, $\tilde{D}$ or $\tilde{E}$. It is not too hard (although it is tedious) to construct such representations whose dimension vector is the imaginary root of the underlying diagram, and prove that they are indecomposable.
For instance, in type $\tilde{A}$, one gets (for any orientation, not just the one pictured)
$$ \begin{array}{ccccccccccc} k[t] & \xrightarrow{t} & k[t] & \xrightarrow{1} & k[t] & \xrightarrow{1} & \ldots & \xrightarrow{1} & k[t] \\ & \searrow_1 & & & & & & \nearrow^1 & \\ & & k[t] & \xrightarrow{1} & \ldots & \xrightarrow{1} & k[t] & & \end{array} $$ and in type $\tilde{D}$ for a certain orientation, one gets $$ \begin{array}{ccccccccccc} k[t] & \xrightarrow{(1,0)^T} & k[t]^2 & \xrightarrow{id} & \ldots & \xrightarrow{id} & k[t]^2 & \xrightarrow{(1,1)} & k[t] \\ & \nearrow_{(0,1)^T} & & & & & & \searrow^{(1,t)} & \\ k[t] & & & & & & & & k[t]. \end{array} $$
I'll add two remarks:
If you don't want to use Gabriel's theorem, then the above still proves the existence of one-parameter families of indecomposable modules for quivers which are not orientations of Dynkin diagrams.
If the field is finite, then one-parameter families are finite as well. To prove that the non-Dynkin quivers admit infinitely many isomorphism classes of indecomposable representations, one needs to consider integer multiples of the dimension vectors used in the last step.