I think there is an obstruction called Van Kampen's obstruction to embedding an $n$-complex into $\mathbb{R}^{2n}$. If you could embed the $n$-complex into $\mathbb{R}^{2n}$ then it has a thickening
to a manifold, and that manifold has an intersection pairing in dimension $n$. The obstruction is derived from this pairing. If the complex embeds the obstruction vanishes. For complexes of dimension greater than 2 this suffices, it is not a complete invariant for two complexes. This is worked out in a paper of Krushkal, Freedman, and Teichner and then an example of imcompleteness of the invariant is in a paper of Krushkal. The example given by FKT is of the two skeleton of the six simplex.

Of course that example has more than one vertex. If you could embed one vertex $2$-complexes in $\mathbb{R}^4$ then some weird stuff would be happening in that example as you collapse a maximal tree in the one skeleton.

The weird thing is that the one skeleton becomes planar. There is a folk theorem that the
classification of four manifolds is undecidable because any finitely presented group is the fundamental group of a four manifold. The proof is to embed a two complex with one vertex,
whose fundamental group is the chosen finitely presented group into $\mathbb{R}^4$ and take a regular neighborhood. Its gotta be Stallings. :)

I would guess that the two questions, one vertex, or planar $1$-skeleton are equivalent.