Let $M, N$ be isomorphic algebraically closed fields over the prime field $k$. For simplicity we assume that $k \subseteq M, N$. Say $\sigma : M \mapsto N$ is our isomorphism. Notice that $\sigma$ preserves the prime field $k$. Moreover, the image of a transcendence basis of $M$ over $k$ is a transcendence basis of $N$ over $k$ (see proof bellow). Hence, $M$ isomorphic to $N$ implies that $M$ and $N$ have the same transcendence degree.
proof
Let $t_1, \dots, t_n$ be a transcendence basis of $M$ over $k$.
First, we check that $\sigma(t_1), \dots, \sigma(t_n)$ is algebraically independent over $k$ :
If not, there is a non zero polynomial $P \in k[X_1, \dots, X_n]$ such that $P(\sigma(t_1), \dots, \sigma(t_n) ) = 0$. But since $P$ has coefficients in $k$, this implies that $P(t_1, \dots, t_n) = 0$, a contradiction!
Then, we show that any element in $N$ is algebraic over $\sigma(t_1), \dots, \sigma(t_n)$ :
Let $b \in N$. The element $a := \sigma^{-1}(b)$ is algebraic over $t_1, \dots, t_n$ i.e. there is a polynomial $P(X) \neq 0$ with coefficients in $k(t_1, \dots, t_n)$ such that $P(a) = 0$.
Write $P^\sigma$ for the polynomial with coefficients in $k(\sigma(t_1), \dots, \sigma(t_n))$ obtained by applying $\sigma$ to all coefficients of $P$. We have $P^\sigma \neq 0$ and $P^\sigma(b) = P^\sigma(\sigma(a)) = \sigma(P(a)) = 0$ : $b$ is algebraic over $\sigma(t_1), \dots, \sigma(t_n)$, whence the claim. $\square$
For the other implication, consider two algebraically closed fields over the prime field $k$ having the same transcendence degree $n$. Let $t_1, \dots, t_n$ (resp. $u_1,\dots, u_n$) be a transcendence basis of $M$ (resp. $N$). Consider the isomorphism $\tau : k(t_1, \dots, t_n) \mapsto k(u_1, \dots, u_n)$ sending $t_i$ to $u_i$ for all $i$. This isomorphism extends to an isomorphism $\sigma : M \mapsto N$ by successive adjunction of algebraic elements.
