If we take the definition of a variety as a reduced integral scheme of finite type over an algebraically closed field $k$, then a variety is in particular a scheme over $k$, so is a scheme $X$ with a morphism $X \to Spec(k)$. As I'm understanding it, more emphasis should be put on the morphism instead of the underlying scheme since we can have non-isomorphic schemes over a base $S$ say $\pi,\eta : X \to S$. Then even if we have a finitely generated reduced $k$-algebra $A$, can we still have non-isomorphic $k$-schemes $\eta, \pi : Spec (A) \to Spec (k)$? It seems to me like if this were the case it would be unexpected, since there is just one $k$ variety called Spec$(A)$, but I don't really have anything to base this on other than the fact that the relative point of view is really never applied to varieties.
This is equivalent to putting two non-isomorphic $k$-algebra structures on $A$, where $k$ is algebraically closed, so I the question I'm really interested in is can we have a ring $A$ and two monomorphisms $i_1,i_2 :k \to A$ so that there is no automorphism $\varphi$ of $A$ with $\varphi \circ i_2 = i_1$?
Here is a non-affine example, although I think (but haven't checked) that it produces an affine example after removing a point: take an elliptic curve $E$ over $k$ whose $j$-invariant $j(E)$ is moved by some automorphism $g : k \to k$ of $k$. Then applying $g$ to the coefficients of a polynomial defining $E$ produces a new elliptic curve $gE$ with $j$-invariant $j(gE) = g j(E) \neq j(E)$. So $E$ and $gE$ are not isomorphic as varieties over $k$, but they are isomorphic as schemes.
In the setting of varieties the point of the structure morphism to $\text{Spec } k$ is mostly to force morphisms to be $k$-linear. Without it you just get the wrong notion of morphism of varieties over $k$.