I have found the following fact stated in a number of places: 


If $k$ is any field, a connected reductive group $G$ is anisotropic if and only if its only unipotent element is $e$ and $\mathrm{Hom}\_k(G, \mathrm{G}\_m)$ is trivial.


For instance, this appears in Section 3.4 of Springer's Corvallis article. However, I have been unable to track down a reference for a proof of this result.



Anisotropic algebraic groups have no unipotent elements
What you have asked for is indeed a fact in characteristic zero. See Corollary 8.5 of Borel-Tits [(Publ. IHES at Numdam, unrestricted access)](http://www.numdam.org/item?id=PMIHES_1965__27__55_0). 


However, it is false in positive characteristic. See also section 4 of the same article (where other fields are considered). Indeed, the group $\mathrm{PGL}\_1(D)$ over a field $k$ of positive characteristic $p$, $D$ a central division algebra over $k$ of degree $p$ , can have "bad" unipotent elements coming from purely inseparable extensions of $k$ lying in $D$. What is true is that $G$ is anisotropic iff it does not have a proper parabolic subgroup defined over $k$.


In any case, this article by Borel and Tits is the standard reference for reductive groups over arbitrary fields.


