You can have plenty of other examples. Take for example a smooth cubic and a curve of degree 3n with 9 points of multiplicity n, with n>1. Both are of geometric genus 1 and you can choose the two curves to be birational. However, no birational map of the plane will send one curve on the other one.

Another way to see that a nodal rational curve of degree $\ge 6$ cannot be sent to a line by a birational map of the plane is to see that it cannot be contracted by a birational map of the plane. Each birational map of the plane decomposes into blow-ups and contractions of smooth $(-1)$-curves (curves isomorphic to $\mathbb{P}^1$ and of self-intersection $-1$). If you want to contract the curve, you have to blow-up all nodes. The number of nodes is $(d-2)\cdot (d-1)/2$, and the self-intersection of the curve on $\mathbb{P}^2$ is $d^2$. After blowing-up, the self-intersection decreases by $4$ for each double point, so becomes $d^2-2\cdot (d-1)(d-2)=6d-d^2+4$. If $d\ge 6$, this number is $\le -2$, so the curve is not contractible. If $d\le 5$, the curve is contractible and in fact one can check easily that it can be sent onto a line by a birational map of the plane.

Similarly, one sees that a nodal rational curve of degree $d\ge 6$ cannot be sent onto a nodal rational curve of degree $d'\ge 6$ when $d\not=d'$.

Similar arguments work with curves of genus $1$. See for example Proposition 3.3.3 of "On birational transformations of pairs in the complex plane", J. Blanc, I. Pan, T. Vust, Geom. Dedicata 139 (2009), 57-73.