The subring $R$ is a finitely generated $k[x,y]$-module since, as an ideal it is $(x)$.
However, as a $k$-algebra it cannot be finitely generated. Take $(f_1,...,f_r)$ to be a finite set in $R$ and set $S$ the algebra generated by $(f_1,...,f_r)$. Set $d_i$ the degree of $f_i$ in $x$. Now we are interested in the degree $1$ in $x$ part of elements of $S$. Since we have $\deg_x(f_if_j)>1$ any degree $1$ in $x$ part of elements of $S$ will come from :
Where $\lambda_i\in k$. Now project $S$ in $R/(x^2)$, we see (from the remark above) that the projection of $S$ on $R/(x^2)$ will be (as a $k$-vectorial space) of dimension $\leq r$. However it is clear that $R/(x^2)$ is (as a $k$-vectorial space) isomorphic to $k[y]$ hence infinite dimensional. So the projection of $S$ doesn't fill the whole space $R/(x^2)$, in particular $S\neq R$.
Edit : Based on Mariano's comment. When one talks about ring some (like me) might not ask for the ring to contain a unity and others (like Mariano) ask for a unity on the ring (most common convention). In the answer above, it is not supposed to contain one. If we want $R$ to contain the unity then, as $k$ vectorial space we have $R:=k\oplus (x)$. Since any element in $k\subseteq R$ is of degree $0$ the rest of the proof remains unchanged. The idea is still the following the degree $1$ part in $x$ $R/(x^2)$ is still isomorphic to $k[y]$ whereas the degree $1$ part of $S/(x^2)$ will be given by a $k$-linear combination of the degree $1$ of the generators of $f$. This gives a finite dimensional space which cannot coincide with the degree $1$ part of $R/(x^2)$.