To prove the equivalence of the statements:

a) $A$ is a field;
b) $M$ is not empty and the smallest element in $M$ is $2^n + 1$,

we will proceed as follows:

1. **Prove that if $A$ is a field, then $M$ is not empty and the smallest element in $M$ is $2^n + 1$:**

- Since $A$ is a field with $2^n + 1$ elements, it is a finite field. In a finite field $\mathbb{F}_q$ with $q$ elements, every element $x$ satisfies $x^q = x$.
- Here, $q = 2^n + 1$. Therefore, for every $x \in A$, we have $x^{2^n + 1} = x$.
- This implies $2^n + 1 \in M$.
- To show that $2^n + 1$ is the smallest element in $M$, assume there exists some $k < 2^n + 1$ such that $x^k = x$ for all $x \in A$.
- Since $A$ is a field, the multiplicative group $A^*$ (the set of nonzero elements of $A$) is cyclic of order $2^n$. Let $g$ be a generator of $A^*$.
- Then $g^{2^n} = 1$. If $g^k = g$, then $g^{k-1} = 1$. This implies $k-1$ must be a multiple of $2^n$, i.e., $k-1 = m \cdot 2^n$ for some integer $m$.
- Since $k < 2^n + 1$, we have $k-1 < 2^n$, which implies $m = 0$ and $k = 1$, which is not possible since $k \geq 2$.
- Therefore, $2^n + 1$ is the smallest element in $M$.

2. **Prove that if $M$ is not empty and the smallest element in $M$ is $2^n + 1$, then $A$ is a field:**

- Assume $M$ is not empty and the smallest element in $M$ is $2^n + 1$.
- This means for every $x \in A$, $x^{2^n + 1} = x$.
- Consider any prime ideal $\mathfrak{p}$ in $A$. In the quotient ring $A/\mathfrak{p}$, we have $(x + \mathfrak{p})^{2^n + 1} = x + \mathfrak{p}$ for all $x \in A$.
- This implies $y^{2^n + 1} = y$ for all $y \in A/\mathfrak{p}$.
- Since $2^n + 1$ is the smallest element in $M$, the order of the multiplicative group of $A/\mathfrak{p}$ must divide $2^n$. Hence, $A/\mathfrak{p}$ is a finite field with $2^m + 1$ elements for some $m \leq n$.
- If $A$ were not a field, it would have a nontrivial prime ideal $\mathfrak{p}$. The quotient $A/\mathfrak{p}$ would be a field with fewer than $2^n + 1$ elements, contradicting the minimality of $2^n + 1$ in $M$.
- Therefore, $A$ must be a field.

Thus, we have shown that the statements are equivalent.

$\blacksquare$