Does the category of affine $k$-varieties have finite products? We use the definitions of this question.
Let $\mathrm{Aff}(k)$ be the category of affine $k$-varieties.
Let $X, Y$ be objects of $\mathrm{Aff}(k)$.
Does the product $X\times Y$ exist in $\mathrm{Aff}(k)$?
 A: There is antiequivalence of categories between affine schemes and rings,  sending a scheme $X=Spec(A)$ to its ring of global functions $A=\Gamma(X, \mathcal O_X)$.
In this correspondence the product $Spec(A)\times Spec(B)$ of two schemes correspond to the tensor product $A\otimes B$ of their rings.
You can "restrict" this antiequivalence to $k$- schemes and thus obtain an antiequivalence between affine $k$-schemes and $k$-algebras sending the product (over $Spec(k)$ ) $X\times_k Y$ of two such $k$-schemes $X,Y$ to the tensor product algebra $\Gamma(X, \mathcal O_X)\otimes _k \Gamma(X, \mathcal O_Y)$.  
And now the fun begins!  
Given that  $X$ and $Y$ have some property, their product may or may not have that property.
For example the product of two integral $k$-schemes is an integral scheme if $k$ is algebraically closed.
For more general $k$ this result  fails: for example over $\mathbb R$ we have $Spec(\mathbb C)\times_\mathbb R Spec(\mathbb C)=Spec(\mathbb C )\sqcup Spec(\mathbb C)$, which is reducible, hence not integral although its factors are integral . (See also the last section of this post).
And if $k$ is an imperfect field of characteristic $p$ with a purely inseparable extension $K=k(a)$ satisfying $a\notin k$ but $a^p \in k$, the scheme $Spec(K)\times_k Spec(K)$ is not reduced and hence not integral although its factors are integral .
[Non reducedness follows from:
  $\sqrt [p]{a}\otimes 1-1\otimes \sqrt [p]{a} \neq 0$ but $(\sqrt [p]{a}\otimes 1-1\otimes \sqrt [p]{a} )^p= a\otimes 1-1\otimes a =0$ ]    
A detailed study of these questions can be found in EGA Chapter IV (Seconde Partie) §4, in which  field theory plays a crucial role.  
An ideological position.
Your linked post refers to the language of Weil's Foundations.
Like most contemporary algebraic geometers I consider that Grothendieck's scheme theory has relegated Weil's theory to an interesting episode  in the history of algebraic geometry but has no place in contemporary mathematics, so that I don't want to have anything to do with that language.
The answer to Makoto's ACTUAL question!
Surprisingly, the full subcategory of schemes with objects reduced schemes has finite products: the product of two reduced schemes $X,Y$ is the reduction  $(X\times Y)_{red} $  of their product $X\times Y $ in the category of schemes .
Essentially, this is because the morphism $X\to X\times Y$  factors through $(X\times Y)_{red} $ if $X$ is reduced and similarly for $Y$ .
The same result holds for reduced finitely generated  $k$-schemes.  
An intriguing example
Consider the purely inseparable extension $K=k(a)$ above ($a\notin k, a^p\in k$).
The coproduct $K\sqcup K$ in the category of $k$-algebras is $K\otimes_k K \cong K[S]/(S-a)^p$, a non-reduced $k$-algebra.
However the coproduct in the category of reduced $k$-algebras also exists and is $K\sqcup'K=K$ (with both morphisms $K\to K\sqcup'K=K$ being the identity).
The reason of this enigmatic result is that given a reduced $k$-algebra $L$, there is at most one morphism of $k$- algebras $K\to L$, and this morphism must send $a\in K$ to the unique $p$-th root $l\in L$  of $a^p\in k$ .
By taking $Spec$'s, you will get an intriguing example of product in the category of reduced affine $Spec (k)$-schemes.
