As there seem to be some differing opinions in the comments as to whether all
irreducible representations are finite-dimensional let me give the argument I
had in mind. A module over the path algebra is the same thing as a
representation of the quiver, which in this case means two vector spaces $U$ and
$V$ and linear maps $e\colon U\rightarrow V$ and $f\colon V\rightarrow U$. We
introduce also $S:=fe\colon U\rightarrow U$ and $T:=ef\colon V\rightarrow
V$. Assume now that $(U,V,e,f)$ is irreducible and pick a non-zero $u\in U$ and
let $W$ be the subrepresentation generated by $u$. It
is clear that the $U$-part of $W$ is the $k[S]$-submodule generated by $u$ so by
irreducibility we have that $U=k[S]u$. Now, do the same argument for $Su$ giving
us also that $U=k[S]Su$. In particular there is a polynomial $p(S)$ such that
$u=p(S)Su$, i.e., $q(S)u=0$ where $q(S)=p(S)S-1$ which in particular is non-zero
so that $U=k[S]u$ is finite dimensional. By symmetry the same argument applies
to $V$ so the module is finite dimensional.

Note, that slightly extending this also gives us a classification of the finite
dimensional modules. In particular there are enough of them to make the Jacobson
radical be equal to $0$.

**Addendum**: Rather than mixing together several steps it is probably better to divide it up: First show that if $(U,V,e,f)$ is irreducible then $U$ is irreducible as $k[S]$-module and then use the classification of simple $k[S]$-module.

Also I feel that the path algebra is something of a red herring. Representations of a quiver are clearly modules over a ringoid (aka ring with several objects). A ringoid with a finite number of objects is Morita equivalent to a ring (as each object gives rise to a compact projective which collectively are faithful, giving a compact faithful projective module as their sum). However, passing to the endomorphism ring of that projective just hides some extra structure that you started with which seems silly.