Question: "How can we show the opposite direction."
Answer: Assume $k \subseteq l:=k\{e_1,..,e_n\}$ is finite and let $K$ be the algebraic closure of $k$ with $L:=K\otimes_k l$. There is an inclusion
$$l\otimes_k l \subseteq L \otimes_k L$$
and isomorphisms $l\otimes_k l:=A \cong k\{e_i\otimes e_j\}$ and $L\otimes_K L:=B \cong K\{e_i\otimes e_j\}$. Hence $l\otimes_k l$ is a finite dimensional $k$-algebra and $L\otimes_k L$ is a finite dimensional $K$-algebra. If $I/I^2\cong \Omega^1_{l/k}=0$ it follows $\Omega^1_{B/K}=0$. If $nil(A) \neq (0)$ there is a maximal ideal $(0)\neq m \subseteq B$ with $B/m \cong K$. There is an isomorphism
$$\Omega^1_{B/K}\otimes_B B/m \cong m/m^2=0$$
hence $m^n=m$ for all $n \geq 2$. Since $dim_K(B)<\infty$ it follows $krdim(B)=0$ hence there is a direct sum decomposition
$$B\cong B_1\oplus \cdots \oplus B_d$$
with $B_i$ artinian $K$-algebra with maximal ideal $m_i \subseteq B_i$. Since $\Omega^1_{B/K}=0$ it follows $\Omega^1_{B_i/K}=0$ hence
$$\Omega^1_{B_i/K}\otimes_{B_i} B_i/m_i \cong m_i/m_i^{2}=0$$
hence $m_i^n=m_i$ for all $n\geq 2$. Since $dim_K(B_i)< \infty$ there is an $N_i$ with $(0)=m_i^{N_i}=m_i=0$. Hence $B_i$ is a field and hence $L\otimes_k L$ is a direct sum of fields, hence $L\otimes_k L$ is reduced. There is an inclusion
$$l\otimes_k l \subseteq L\otimes_K L$$
and it follows $l\otimes_k l$ is reduced - a contradiction, hence $nil(A)=(0)$.
Hint: $K\otimes_k \Omega^1_{l/k} \cong \Omega^1_{L/K}$ and $\Omega^1_{B/K} \cong L\otimes_K \Omega^1_{L/K}\oplus \Omega^1_{L/K} \otimes_K L$. Since $B:=L\otimes_K L$ is a finite dimensional non-reduced algebra, $K$-algebra there is a maximal ideal $m \subseteq B$ and since $K$ is algebraically closed it follows $B/m \cong K$.
Example: If $k:=\mathbb{Q}$ and $K:=k(\sqrt{2})$ it follows $k \subseteq K$ is separable and $\Omega^1_{K/k}=0$. You may check explicitly that
$$K\otimes_k K \cong K \oplus K$$
is a direct sum of fields, hence $K\otimes_k K$ is reduced.
Conversely: If $A$ is reduced we may argue as follows: There is a direct sum decomposition
$$A \cong A_1 \oplus \cdots \oplus A_d$$
with $A_i$ artinian, and since $A$ is reduced it follows $A_i$ is a field for all $i$. Since $I\subseteq A$ is a maximal ideal it follows $I:=A_1\oplus \cdots \oplus (0) \oplus \cdots \oplus A_d$ and hence for any element $a\in I$ it follows the element $xa=a \in I^2$ where
$$x:=(1,..,1,0,1,..,1)\in I.$$
Hence $I^2=I$. We have proved:
Lemma: Let $k \subseteq l$ be a finite extension of fields. It follows $\Omega^1_{l/k}=0$ iff $l\otimes_k l$ is reduced.
Best Answer
Let's prove the following: If $E/F$ is an algebraic extension of fields such that $E \otimes_F E$ is a domain, then $E = F$.
So let $\alpha \in E$. Then $F(\alpha) \otimes_F F(\alpha)$ embeds into $E \otimes_F E$, which is therefore also a domain.
Let $f \in F[T]$ be the minimal polynomial of $\alpha$. Then we compute the tensor product as $$F(\alpha) \otimes_F F(\alpha) \cong F(\alpha) \otimes_F F[T]/ \langle f \rangle \cong F(\alpha)[T] / \langle f \rangle.$$ Now, this ring is a domain. This means that $f = 0$ (which is not the case) or that $f$ is irreducible over $F(\alpha)$. But $f$ has a root in $F(\alpha)$, namely $\alpha$. It follows $\deg(f)=1$, i.e. $\alpha \in F$.
Notice that the claim is not true for non-algebraic extensions. For example, $F(T) \otimes_F F(T)$ is a domain (not a field!), since it is a localization of $F[T] \otimes_F F(T) \cong F(T)[T']$, which is a domain.