A node on a curve is a singular point that locally looks like the intersection of two lines. I think the precise way to say this is that $p \in X$ is a (closed?) point on a scheme $X$ (of finite type over a field $k$?), then the completion of the local ring at $p$, $\widehat{\mathcal{O}}\_{X,p}$ should be isomorphic to $k[[x,y]]/(xy)$ (the completion for the intersection of two lines).


The first question is whether this is correct. 


The second question is whether you get the n-dimensional version of a node by requiring
the completion of the local ring to look like intersection of $n+1$ coordinate planes: $k[[x\_0,x\_1,...,x\_n]]/(x\_0 x\_1\cdots x\_n)$.


The third question is: what exactly does it mean for such a singularity to be isolated? This 
is easy to imagine over $\mathbb{C}$ (there is analytic neighborhood containing no other singularity), but how to say this in something like the Zariski topology? 



Higher dimensional nodes
**[Edit: Over an algebraically closed field]** a node should be an isolated hypersurface singularity whose (projectivized) tangent cone is a nondegenerate quadric. This means that in local coordinates the equation has no linear part, and the quadratic part is nondegenerate.


For a curve your definition is equivalent to this one so it is correct. In higer dimensions it is very different. In fact your hypersurface has a non-isolated singularity (for instance, all points in the coordinate lines are singular).


For the last question, the singular points of an algebraic set (or variety, or scheme) are a Zariski closed subset, which splits in irreducible components. A singular point is isolated if it is one of the irreducible components (and it is not embedded, ie, it does not belong to any other irreducible component, but this has sense only in the scheme-theoretic case). 


For the **non algebraically closed field** case, see BCnrd's comment and reference!


