As we know every finite boolean ring in which $1≠0$ is isomorphic to, $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \cdots \times \mathbb{Z}_2$.
So, is every infinite boolean ring is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \cdots\ $(infinite copies)?
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No. For instance, if $I$ is any set, then any subring of the product $\mathbb{Z}_2^I$ will also be a Boolean ring, and such subrings usually are not themselves products of copies of $\mathbb{Z}_2$. For instance, any infinite product of copies of $\mathbb{Z}_2$ must be uncountable, but you can take the subring of $\mathbb{Z}_2^{\mathbb{N}}$ generated by a countably infinite subset to get a subring which is countably infinite. Such a subring cannot be isomorphic to any product of copies of $\mathbb{Z}_2$, since it is infinite but not uncountable.
However, every Boolean ring has this form: every Boolean ring is isomorphic to a subring of $\mathbb{Z}_2^I$ for some set $I$. This is a somewhat hard theorem (a weak form of Stone's representation theorem).
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Note that the Boolean algebra corresponding to a ring of the form $\prod_{i\in I}\mathbb{Z}/2\mathbb{Z}$ will have atoms - that is, nonzero elements which don't properly bound any nonzero elements (think about the element corresponding to $\delta_i$ for $i\in I$ ...). So any atomless Boolean algebra will yield a Boolean ring which is not of that form.
No, there are many infinite Boolean algebras (equivalent to a Boolean ring), not only the powers of $\{0,1\}$. E.g the Borel sets in $\mathbb{R}$.