I claim that no infinite dimensional algebra $A$ over $k$ satisfies 2. I am indebted to the comment of @UriyaFirst for the idea of the main step in the proof.
First I claim we may assume that $k$ is algebraically closed. Indeed, if $[A:k]=\infty$ and satisfies 2, then for an algebraic closure $\overline k$ of $k$ we have that $A'=A\otimes_k \overline k$ is infinite dimensional over $\overline k$ and satisfies 2 over $\overline k$ by transitivity of extension of scalars.
So assume $k$ is algebraically closed and $[A:k]=\infty$. We show that $A$ does not satisfy $2$. If $A$ is not semisimple, then it fails 2, so we may assume that $A$ is a direct product of matrix algebras over division algebras and at least one of these division algebras $D$ is infinite dimensional over $k$. Since extension of scalars commutes with direct product and matrix algebras, it suffices to show that $D\otimes_k K$ is not semisimple for some field extension $K/k$.
The key observation is that if $F/k$ is an infinite extension, then $F\otimes_k F$ is either not Artinian or not Noetherian. If $F/k$ is not finitely generated, then $F\otimes_k F$ is not Noetherian by Theorem 11 of P. Vámos, On the minimal prime ideals of a tensor product of two fields, Mathematical Proceedings of the Cambridge Philosophical Society, 84 (1978), pp. 25-35. If $F/k$ is finitely generated but infinite dimensional then it has positive transcendence degree. The Krull dimension of $F\otimes_k F$ is the transcendence degree of $F/k$, by Rodney Y. Sharp, The Dimension of the Tensor Product of Two Field Extensions, Bulletin of the London Mathematical Society 9 Issue 1 (1977) pp 42–48. Thus $F\otimes_k F$ is not Artinian in this case.
There are two cases. If $[Z(D):k]=\infty$, then $D\otimes_k Z(D)$ has center $Z(D)\otimes_k Z(D)$ by general facts on centers of tensor products over a field, and so the center of $D\otimes_k Z(D)$ is not a finite direct product of fields by the above observation (since it fails either the Artinian or Noetherian condition). But the center of a semisimple ring is a direct product of fields by Wedderburn-Artin, contradiction.
So assume $[Z(D):k]<\infty$. Then $Z(D)=k$ because $k$ is algebraically closed. Let $\alpha\in D\setminus k$. Then $\alpha$ is transcendental over $Z(D)=k$ and so $k\leq K=k(\alpha)$ with $K/k$ a purely transcendental extension. Now using @UriyaFirst comment (but with a descending chain argument), we have that $K\otimes_k K$ is not Artinian by the above and so has an strictly descending chain of ideals $I_1\supsetneq I_2\supsetneq \cdots$. Now $D\otimes_k K$ is a free right $K\otimes_k K$-module (since $D$ is a free $K$-module) and so the $(D\otimes_k K) I_k$ form a strictly descending chain of left ideals in $D\otimes_k K$. Thus $D\otimes_k K$ is not left Artinian and hence not semisimple.