Let $k$ be an algebraically closed field of characteristic $\neq 2$ and consider the affine varieties $X_1 := V(x^2-y^3)$ (the cusp) and $X_2 = V(y^2-x^2 (x+1))$ (the node). I would like to show that $X_1$ and $X_2$ are not isomorphic. This is easy if one knows some basic cohomology or singularity theory; but this exercise appears in an introductory script to AG where only absolute basics about affine varieties have been established. Thus, I'm looking for an elementary proof. But of course I would like to avoid any fiddly computation with polynomials (please don't post these as an answer).
Here is what I've done: There are bijective morphisms $\mathbb{A}^1 \to X_1, t \mapsto (t^3,t^2)$ and $\mathbb{A}^1 \to X_2, t \mapsto (t^2-1,t(t^2-1))$. They induce isomorphisms $\mathbb{A}^1 - \{0\} \cong X_1 - \{(0,0)\}$ (the inverse takes $(x,y) \mapsto x/y$) and $\mathbb{A}^1 - \{\pm 1\} \cong X_2 - \{(0,0)\}$ (the inverse takes $(x,y) \mapsto y/x$). Now any isomorphism $X_1 \cong X_2$ must preserve the origin. The reason is that it is the unique singular point (which can be formulated elementarily, see here). Thus, it would yield an isomorphism $\mathbb{A}^1 - \{0\} \cong \mathbb{A}^1 - \{\pm 1\}$. On coordinate rings, this is an isomorphism $k[t]_t \cong k[t]_{t^2-1}$ of $k$-algebras. It induces an isomorphism on the groups of units $k^* \cdot \langle t \rangle \cong k^* \cdot \langle t+1,t-1 \rangle$, preserving $k^*$, thus an isomorphism of groups $\mathbb{Z} \cong \mathbb{Z} \oplus \mathbb{Z}$, contradiction!
I hope that there is a shorter proof? As I've said, I'm only looking for proofs which are accessible to students which have just started to learn about affine varieties.