Firstly, suppose $n=2^k-1$. Then $\binom{n}{m}=\frac{(2^k-1)(2^k-2)....(2^k-m)}{1*2*......*m}$. For any $a,b$ with $b<k$, $a \equiv -(2^k-a) \mod{2^b}$, so in particular, if one is zero, the other is zero also. Therefore, comparing factors in the numerator and the denominator, all the powers of two will cancel out exactly, leaving an odd binomial coefficient, completing the sufficiency part of the proof. Now, suppose $n=2^k-l$ with $1<l\leq2^{k-1}$. Consider the binomial coefficient $\binom{2^k-l}{2^{k-1}-1}=\frac{(2^k-l)!}{(2^{k-1}-1)!(2^{k-1}-l+1)!}=\frac{(2^k-l)(2^k-l-1)....2^{k-1}}{(2^{k-1}-l+1)!}$ Factoring out, this becomes $\frac{2^{k-1}}{2^{k-1}-l+1}\binom{2^k-l}{2^{k-1}-l}$, which is clearly even by choice of l.

Other method is Lucas theory for the case $ p = 2$ $ n = b_k.2^k + b_{k - 1}.2^{k - 1} + ... + b_1$ $ m = c_{k}2^k + c_{k - 1}2^{k - 1} + .. + c_1$ Where $ b_i,c_i\in\{0,1\}$ Then $ C_n^m\equiv \prod_{i = 1}^k C_{b_i}^{c_i}(\mod 2)$ Imply that $ C_{b_i}^{c_i}\equiv 1 (\mod 2)$ So $ b_1 = b_2 = ... = b_k=1$ imply that $ n = 2^{k + 1} - 1$ This problem was be prove.

The binomial coefficients are of the form \[\binom{n}{i} | 0\le i\le n\] In base $2$, let $n=\sum_{i=0}^{j}2^ia_i$ where $a_j=1$. By Lucas's theorem if there exists an $m$ such that $a_m=0$ then we would have \[\binom{n}{2^m}\equiv \binom{0}{1}\pmod{2}\equiv 0\pmod{2}\] Contradiction. Therefore $a_m=1\forall 0\le m\le j$ and hence $n=\sum_{i=0}^{j}2^i=2^{j+1}-1$. We are done.

If I'm not mistaken, is that not precisely the proof above it, with just $a_i$ replacing $b_i$, and using the same Lucas' theorem?

$\dbinom{n}{r} =\frac{n!}{(n-r!)\times(r!)}$ is odd if and only if $v_2(n!)=v_2(r!)x+v_2(n-r!) \iff \sum_{i=1}^{\infty} [\frac{n}{2^i}]=\sum_{i=1}^{\infty} [\frac{n-r}{2^i}]+\sum_{i=1}^{\infty} [\frac{r}{2^i}] \iff$ $\sum_{i=1}^{\infty} ([\frac{n}{2^i}]-[\frac{n-r}{2^i}]-[\frac{r}{2^i}])=0 \iff \forall i \in N: [\frac{n}{2^i}]=[\frac{n-r}{2^i}]+[\frac{r}{2^i}]$ now assume$\dbinom{n}{r}$ s are all odd it mean $\forall r<n,\forall i\in N : [\frac{n}{2^i}]=[\frac{n-r}{2^i}]+[\frac{r}{2^i}]$ we know there exists $k$ such as that $2^{k-1}\le n <2^k \implies n=2^k-1-s$ for some $s\ge 0 \implies$ now if we put $i=k-1$ and $r=2^{k-1}-1$ $[\frac{n}{2^i}]=[\frac{n-r}{2^i}]+[\frac{r}{2^i}] \implies[\frac{2^k-1-s}{2^{k-1}}]=[\frac{2^{k-1}-s}{2^{k-1}}]+[\frac{2^{k-1}-1}{2^{k-1}}]$ because LHS is equal to one and $[\frac{2^{k-1}-1}{2^{k-1}}]=0 \implies [\frac{2^{k-1}-s}{2^{k-1}}]=1 \implies s=0$ so $n=2^k-1$ we other part is easy, let $n=2^k-1$ we should prove $\forall r<n,\forall i\in N : [\frac{n}{2^i}]=[\frac{n-r}{2^i}]+[\frac{r}{2^i}] \implies$ if $i \ge k$ then both sides will be zero so supose $i<k$ $\forall r<n,\forall i\in N : [\frac{2^k-1}{2^i}]=[\frac{2^k-1-r}{2^i}]+[\frac{r}{2^i}]$ let $r=2^i\times t+r'$ that $2^i>r'\ge 0,t \ge 0 \implies [\frac{2^k-1}{2^i}]=[\frac{2^k-1-2^i\times t-r'}{2^i}]+[\frac{2^i\times t+r'}{2^i}]$ LHS is obviously $2^{k-i}-1$ and for RHS , $[\frac{2^i\times t+r'}{2^i}]=t$ and $[\frac{2^k-1-2^i\times t-r'}{2^i}]=-t+[\frac{2^k-1-r'}{2^i}]$ and it's easy to see $2^{k-i}-1 \le \frac{2^k-1-r'}{2^i} < 2^{k-i}$ so $[\frac{2^k-1}{2^i}]=[\frac{2^k-1-r}{2^i}]+[\frac{r}{2^i}]$

the coefs of $(a+b)^n$ are of the form $\binom{n}{t}$ for some integer $t \le n$ then we have \[\binom{n}{t}=\frac{n!}{t!(n-t)!} \qquad \textbf{which is odd}\]by the parity condition we have that \begin{align*} v_2(n!) & = v_2(t!)+v_2((n-t)!)\\ n-s_2(n)& = n-s_2(t)-s_2(n-t)\\ & \implies s_2(n)=s_2(n-t)+s_2(t) \end{align*}since the last equation holds for $n+1$ different variables we easily obtain that the binary representation of $n$ is equal to $111\cdots 111$ which is obviously in the form $2^k-1$ as desired.