Update: Paul Levy points out in the comments that a reasonable way of defining the nullcone in $\operatorname{Lie}(G) = \mathfrak{g}$ is as the zero set of the homogeneous invariants of positive degree — i.e. of $(k[\mathfrak{g}]_+)^G = (S^+\mathfrak{g}^*)^G$.
But with this definition, my original comment isn't valid. Indeed, if $G = \operatorname{PGL}_2$ then the co-adjoint representation of $G$ on $\mathfrak{g}^*$ has a fixed vector, which is a linear invariant in $(k[\mathfrak{g}]_+)^G$ whose zero locus is — as in Dave Stewart's original answer — the span of the root vectors.
I suppose all my original objection (below) really amounted to was that if $X$ is the affine scheme defined by the $\mathbf{Z}$-algebra $R=\mathbf{Z}[A,B,C]/\langle A^2 + 4BC\rangle$, then for any field $k$ of char. not 2, $X_k$ identifies with the nilpotent variety of $\mathfrak{pgl}_{2,k}$, but if $k$ has char. 2, $X_k$ is not reduced.
Original post: I'm going to write this as an "answer", though I think it mostly amounts to a comment on Dave Stewart's answer.
It is not completely clear to me that the statement "the nilpotent variety of $\mathfrak{pgl}_2$ is reduced" is correct when $p=2$. Well, I suppose that more precisely I mean: it isn't clear that the scheme of nilpotent elements should be viewed as reduced.
Taking a basis $x,y,h$ of $\mathfrak{pgl}_2$ (say, in its 3-dimensional representation), one finds that $ah + bx + cy$ is nilpotent just in case $a^2+4bc=0$. Of course, in char. 2 this amounts to $a^2=0$, which suggests that the scheme of nilpotent elements shouldn't be viewed as a reduced subscheme. (If you don't want to write down the matrices, see e.g. Jantzen "Nilpotent Orbits in Representation Theory" §2.7.)
I do doubt (?) that this nilpotent scheme is isomorphic to the scheme of unipotent elements of $\operatorname{PGL}_2$, but (assuming that doubt is correct — I didn't think too carefully about it) I think the reason is more complicated than the statement "one is reduced and the other isn't".
