Geometry
Start by proving that a Boolean ring is commutative. Next, show that if $R$ is a commutative ring, idempotent elements $e \in R$ are in natural bijection with decompositions
$$R \cong Re \times R(1 - e)$$
of $R$ into a direct product of two rings. Geometrically this says that idempotents in $R$ correspond to ways to disconnect the affine scheme $\text{Spec } R$ into two components: think of an idempotent as a function $\text{Spec } R \to \{ 0, 1 \}$ to see the topological intuition behind this.
So a Boolean ring is a very special sort of ring: it's a commutative ring where every element corresponds to a way to decompose $\text{Spec } R$ into two pieces. The proof that finite Boolean rings are power sets can be done by inducting on this observation, repeatedly disconnecting $\text{Spec } R$ into pieces until you get down to pieces that can't be disconnected any further. This means Boolean rings where the only idempotents are $0$ and $1$ (rings with this property are called connected, as you might expect), and of course $\mathbb{F}_2$ is the unique such ring.
So geometrically a better way to state the classification is that every finite Boolean ring is the ring $X \to \mathbb{F}_2$ of $\mathbb{F}_2$-valued functions on a finite set $X$. In fact something even better is true: the category of finite Boolean rings is equivalent to the opposite of the category of finite sets.
Logic
Boolean rings are the same thing, in a very strong sense, as Boolean algebras (more precisely, they are equivalent as concrete categories). Boolean algebras are supposed to describe how Boolean logic works, and power sets are the most natural models of Boolean logic for the following reason. In the language of Boolean logic, you can interpret a morphism $B \to \mathbb{F}_2$ from a Boolean ring $B$ to $\mathbb{F}_2$ as a consistent assignment of truth values to each element of $B$, thought of as logical propositions. If $B$ is a collection of questions about the state of a system, then such morphisms are possible answers, generated by possible states of the system.
So conversely it's very natural to associate to every $b \in B$ the set of all states in which $b$ is true. If there are "enough" states, then this gives a power set description of $B$.
Commutative algebra
Another approach is the following. Associated to every commutative ring $R$ is a canonical map
$$R \to \prod_P R/P$$
from $R$ to the product of the quotients $R/P$ of $R$ by every prime ideal. (In algebraic geometry terms, we are evaluating elements of $R$ at every point of the prime spectrum.) The kernel of this map is the intersection of the prime ideals of $R$, which is known to be the nilradical. Now, show that
- A Boolean ring has no nontrivial nilpotent elements, and
- If $B$ is a Boolean ring, then so is every quotient of $B$, and if $B$ is also an integral domain, then $B = \mathbb{F}_2$.
From here it follows that every Boolean ring admits a canonical injection
$$B \to \prod_P B/P \cong \prod_P \mathbb{F}_2$$
into a product of copies of $\mathbb{F}_2$ (and in particular it follows that homomorphisms $B \to \mathbb{F}_2$ are in natural bijection with prime ideals of $B$). From here you need some extra argument to show that the above map is also surjective when $B$ is finite. I think you can use the Chinese remainder theorem.
The benefit of this approach is that it generalizes to some related situations. For example, it follows that every commutative ring in which every element satisfies $x^n = x$ for some fixed $n$ (Jacobson famously showed that you can drop "commutative" here) canonically injects into a product of finite fields of order $q$ where $q - 1$ divides $n - 1$, and for finite rings this is a bijection.
Further
To get some context, you can also learn about what infinite Boolean rings are like here. "Finite set" gets replaced by "profinite set."
The ring structure is essentially unique. Write elements as formal sums
$$
(a_0,a_1,a_2,\dotsc)=\sum_n a_n
$$
with $a_n\in I^n$ (and all zero, but a finite number). Since you want the product to be distributive with respect to addition, and
$$
\Bigl(\sum_n a_n\Bigr)+\Bigl(\sum_n b_n\Bigr)=\sum_n (a_n+b_n)
$$
you have to define the product of $a_m$ by $b_n$, where $a_m\in I^m$ and $b_n\in I^n$: since $a_mb_n\in I^{m+n}$, the choice is obvious.
The verification of the ring laws is a bit tedious, but simple.
For the second part, define $e_0=(1,0,0,\dotsc)\in\operatorname{Rees}_R((a))$, $e_1=(0,a,0,\dotsc)$ and note that $(e_1)^k=e_k$, the element having $a^k$ at the $k$-th position and $0$ elsewhere.
The subring generated by $e_0$ is clearly isomorphic to $R$ and the action of $R$ on $\operatorname{Rees}_R((a))$ as module is the same as the action of the subring. You can so define uniquely a ring homomorphism $\varphi\colon R[x]\to\operatorname{Rees}_R((a))$ by $\varphi(r)=re_0$ and $\varphi(x)=e_1$. Explicitly,
$$
\varphi(r_0+r_1x+\dots+r_nx^n)=r_0e_0+r_1e_1+\dots+r_ne_n
$$
This homomorphism is surjective (prove it). Now compute the kernel.
Best Answer
You're almost there. What you showed in the end is that the identity map is not a ring homomorphism from $\mathbb Z$ with $n$ multiplication to $\mathbb Z$ with $m$ multiplication. So indeed this is not an isomorphism. However, a priori there could be some other isomorphism. However, the only group homomorphisms $\mathbb Z \longrightarrow \mathbb Z$ are multiplication by some fixed integer, so the only group isomorphisms are multiplication by the units $\pm 1$. You've shown that the identity map is not a ring isomorphism, so it remains to consider multiplication by $-1$. But since you restricted to $m > 0$, this won't work either. Hence, these rings are not isomorphic.