If $B$ has one or two elements, $|B|$ is a power of $2$. Assume that $|B|$ is a power of $2$ for all Boolean lattices $B$ with $|B|\lt n$ for some $n\gt2$, and consider a Boolean lattice $B'$ with $|B'|=n$. Pick any element $a$ other than $0$ and $1$ in $B'$. Since $B'$ is a complemented lattice, by the lemma, $B'$ is isomorphic to $\downarrow a\times\uparrow a$, and since $a$ is neither $0$ nor $1$ both $\downarrow a$ and $\uparrow a$ have more than one element. Thus they are both Boolean lattices of size less than $n$. The size of $B'$ is the product of their sizes, which by the induction hypothesis are powers of $2$, so $|B'|$ is also a power of $2$. It follows by complete induction that the cardinality of all finite Boolean lattices is a power of $2$.
[Edit in response to comment:]
$\downarrow a$ and $\uparrow a$ are Boolean lattices because they are closed under the lattice operations,
$ (x\lor a)\lor(x\lor b)=x\lor(a\lor b)\;,\\ (x\lor a)\land(x\lor b)=x\lor(a\land b)\;,\\ (x\land a)\lor(x\land b)=x\land(a\lor b)\;,\\ (x\land a)\land(x\land b)=x\land(a\land b)\;, $
they inherit distributivity from $B'$, and they are complemented:
$ (x\lor a)\land 1=x\lor a\;,\\ (x\lor a)\lor x=x\lor a\;,\\ (x\land a)\land x=x\land a\;,\\ (x\land a)\lor 0=x\land a\;, $ $ (x\lor a)\lor(x\lor\bar a)=1\;,\\ (x\lor a)\land(x\lor\bar a)=x\;,\\ (x\land a)\lor(x\land\bar a)=x\;,\\ (x\land a)\land(x\land\bar a)=0\;. $