Equivalent definitions of perfect field Definition 1: Let $k$ be a field. It is called perfect  field, if $\text{char} \ k=0$ or $\text{char} \ k=p$ with $k^p=k$.
Definition 2: $k$ has no properly bigger purely inseparable extension.
These definitions are equivalent.
Proof: $1 \Rightarrow 2$
If $\text{char} \ k=0$ I don't know what to do.
If $\text{char} \ k=p$. Suppose that exists an extension $E$ of $k$ which is properly bigger and is purely inseperable. Take $\alpha\in E$ such that $\alpha\notin k$. Then $\alpha^{p^n}\in k$ for some $n\geq 1$. Since $k=k^p$ then $\alpha^{p^n}=\beta_1^p$ for $\beta_1\in k$. Hence by freshman's dream $(\alpha^{p^{n-1}}-\beta_1)^p=0$  and $\alpha^{p^{n-1}}=\beta_1$ and if we continue in this way we get $\alpha^p=\beta_{n-1}\in k$ and $\beta_{n-1}=\beta_n^p$ for $\beta_n\in k$. Hence $\alpha^p=\beta_n^p$ and $(\alpha-\beta_n)^p=0$ so $\alpha=\beta_n\in k$ which is contradiction because $\alpha \notin k$.
$2 \Rightarrow 1$ Unfortunately I don't know how to prove this direction. I have tried by contradiction but it was useless.
Can anyone show how to prove it, please?
EDIT: An algebraic extension $E$ over $k$ is called purely inseparable if any of the following is true:
1) $[E:k]_s=1$.
2) All elements of $E$ are purely inseparable over $k$.
3) For any $\alpha\in E$, $\text{Irr}(\alpha,k,X)=X^{p^d}-a$ with $a\in k$, $d\geq 0$.
4) $E$ is generated over $k$ by a (possibly infinite) family $\{\alpha_i\}$ of elements all purely inseparable over $k$.
 A: [Original answer below, answer to edited question first]
As definition $2$ does not apply to fields $k$ with $\operatorname{char}k=0$, the two definitions cannot be equivalent. If definition $1$ is considered only for fields of positive characteristic, then the two are equivalent. That is to say; for a field $k$ of positive characteristic $p$, the following two definitions are equivalent:
Definition 1: The field $k$ is called perfect if $k^p=k$.
Definition 2: The field $k$ is called perfect if it has no proper purely inseparable extension.
Your proof of the implication $1\ \Rightarrow\ 2$ is fine, though you might want to emphasize at the very start that you are assuming $k^p=k$. For a cleaner argument, you could work with the minimal $n\geq1$ such that $\alpha^{p^n}\in k$ to arrive at your conclusion more quickly.
For the converse, if $k$ has no purely inseparable extension then $X^{p^d}-a$ is reducible for all $a\in k$ and all $d\geq0$. In particular $X^p-a$ is reducible for every $a\in k$, where $X^p-a=(X-\alpha)^p$ for some $\alpha$ in a splitting field of $X^p-a$. This means $(X-\alpha)^m\in k[x]$ for some $0<m<p$. The coefficient of $X^{m-1}$ is $-m\alpha$ where $-m$ is a unit in $k$, and so we see that $\alpha\in k$. This proves that $k=k^p$.

Here is an outline of the proof of the contrapositive, where $k$ is a field:
If $k$ has a proper purely inseparable extension, then there exists an irreducible inseparable polynomial $f\in k[X]$. Then $f'=0$ so $\operatorname{char}k=p$ for some prime number $p$, and $f(X)=g(X^p)$ for some $g\in k[X]$. If $k^p=k$ then all coefficients of $g$ are $p$-th powers, so $f(X)=g(X^p)$ is a $p$-th power, contradicting the irreducibility of $f$. Hence $k^p\neq k$.
For the converse; if $\operatorname{char}k=p$ and $k^p\neq k$ then for every $a\in k-k^p$ the polynomial $f=x^p-a\in k[x]$ is irreducible and inseparable.

Note that a field $k$ has an inseparable extension if and only if there exists an irreducible inseparable polynomial $f\in k[x]$. This brings your definition $2$ in accordance  with definition $1$ of these notes.
A: There is no inseparable element over a field of characteristic $0$. So we can reduce to fields of characteristic $p>0$.
Suppose $k^p=k$ and let $E$ be a purely inseparable extension field of $k$. Take $a\in E\setminus k$ such that its minimal polynomial has minimal degree, say $X^{p^e}-b$, with $b\in k$ and $e\ge1$; since $k^p=k$, we get that $b=c^p$ and this is a contradiction, because
$$
X^{p^e}-b=(X^{p^{e-1}}-c)^p
$$
is reducible.
Conversely, suppose $b\in k\setminus k^p$. Then the polynomial has multiple roots by the derivative test; let $a$ be a multiple root in some extension field $E$, so $X^p-b=(X-a)^mf(X)$, with $m>1$ the multiplicity of the root $a$, so $f(a)\ne0$. The derivative is then
$$
0=m(X-a)^{m-1}f(X)+(X-a)^mf'(X)=(X-a)^{m-1}(mf(X)+(X-a)f'(X))
$$
As $m>1$, we conclude that $mf(X)+(X-a)f'(X)=0$. If $m<p$, we can evaluate at $a$ and get $mf(a)=0$, contradicting $f(a)\ne0$. Hence $m=p$, which means $X^p-b=(X-a)^p$. A factorization of $X^p-b$ over $k[X]$ must then have the form $(X-a)^{n_1}(X-a)^{n_2}\dots(X-a)^{n_r}$. Take $n=\min\{n_1,\dots,n_r\}$ and suppose $n<p$ (that is, that $X^p-b$ is not irreducible). Then
$$
(X-a)^n=\sum_{j=0}^n \binom{n}{j}(-1)^ja^{n-j}X^j\in k[X]
$$
Since $n<p$, $\binom{n}{n-1}(-1)^j\ne0$ in $k$, so $a\in k$. Contradiction to $b\notin k^p$.
The extension $k(a)$ is inseparable and has degree $p$, so it is purely inseparable.
