Let $A$, $B$ be two reduced $k$-algebras. Then if an element of the form $$\sum a_{i}\otimes b_{j}$$ is nilpotent, we can compose it with any $k$-homomorphism $f$ from $A$ to $k$ to get a homomorphism from $A\otimes B$ to $B$. This map's image is $$\sum f(a_{i})\otimes b_{j}$$must be nilpotent as well. So if we assume $b_{j}$ is linearly independent in the first place, we may reason that $f(a_{i})=0$. This showed that $a_{i}$ are in the Jacobson radical of $A$. However, we only know $\mathrm{Nil}(A)=0$, and it is not clear to me how to pass from the Jacobson radical to the nilradical without using some commutative algebra machinery.
Here is what I think that might work. If $A$ is of finite type, then by Nullstellensatz we can assert the above is true. But if $A$ is not of finite type, then I can write $A=k[x_{1},...,x_{n}][X]$, where $n$ is the transcendence degree of $k(A)$ and $k[X]$ is finite over $k$. Then I should be able to pass from $k[X]$ to $A$ by inductive argument on the transcendental degree of $k(A)$, since prime ideals are maximal in a UFD (I am working with $k[x]$ by introducing $[x]$).
The proof I have for the case of finite type is not satisfying. Given $I\subset k[X]$, using Nullstellensatz one can show that $$I_{X}V_{X}[I]=\sqrt{I}$$ where $V_{X}[I]$ is the set of all points in $X$ on which $I$ vanishes, and $I_{X}(V_{X}[I])$ be the ideal in $k[X]$ which vanishes on $V_{X}[I]$. Granting this, let $I=(a_{i})$, then we want to show $I=0$, which is $V_{X}[I]=X$. This follows since if $U\subset W$, then $V_{X}[U]\supset V_{X}[W]$. Since $V_{X}[I]\supset V_{X}[m_{x}]$ for all maximal ideal corresponding to points, it must be $X$. This proof is unnecessarily winding by using the ideal-variety correspondence. Ideally I should be able to construct a pure algebraic proof.
The thing not clear to me is whether the above strategy really works. Namely I am not sure if I can really induct on the transcendental degree of $k(A)$. To make matters simple assume we showed the statement holds for $A$ with transdental degree $<n$. For $A$ with transdental degree $n$ we can thus write it as $k[A'][x]$. Then any maximal ideal $m'$ of $k[A'][x]$ must correspond to a maximal ideal $m=m'\cap k[A']$ in $k[A']$. But it seems not possible to show that any prime ideal is maximal. Indeed considering the simple example of $k[x,y]$, where $(x)$ is a prime ideal but not maximal since $k[x,y]/(x)=k[y]$ is not a field. The problem is in the inducting process the base ring is not neccessarily a field, and so this strategy cannot work.
I am thinking maybe I can show $m'=m[f(x)]$ with $f(x)$ some irreducible polynomial in $k[A'][x]$. Because we know $k[A'][x]$ is noetherian by Hilbert's basis theorem, so in particular $m'$ must have a finite generating set. And we can reduce the number of irreducible polynomial to 1 since $m'$ is a maximal ideal. But I still do not know how to use this to finish the proof.
