Here is a variation which transforms the left-hand side into the right-hand side. In order to do so we use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write e.g.\begin{align*}[z^k](1+z)^n=\binom{n}{k}\end{align*}
We obtain for $0\leq m\leq n$:
\begin{align*}\sum_{i=0}^n\binom{n}{2i+1}\binom{i}{m}&=\sum_{i= m}^\infty\binom{n}{n-2i-1}\binom{i}{m}\tag{1}\\&=\sum_{i= 0}^\infty\binom{n}{n-2i-2m-1}\binom{i+m}{m}\tag{2}\\&=\sum_{i= 0}^\infty\binom{n}{n-2i-2m-1}\binom{-m-1}{i}(-1)^i\tag{3}\\&=\sum_{i= 0}^\infty[z^{n-2i-2m-1}](1+z)^n[u^i](1-u)^{-m-1}\tag{4}\\&=[z^{n-2m-1}](1+z)^n\sum_{i= 0}^\infty z^{2i}[u^i](1-u)^{-m-1}\tag{5}\\&=[z^{n-2m-1}](1+z)^n(1-z^2)^{-m-1}\tag{6}\\&=[z^{n-2m-1}](1+z)^{n-m-1}(1-z)^{-m-1}\tag{7}\\&=\sum_{i=0}^{n-m-1}\binom{n-m-1}{i}[z^{n-2m-1-i}]\sum_{j= 0}^\infty\binom{j+m}{j}\tag{8}\\&=\sum_{i=0}^{n-m-1}\binom{n-m-1}{n-m-1-i}\binom{n-m-1-i}{n-2m-1-i}\tag{9}\\&=\binom{n-m-1}{m}\sum_{i=0}^{n-m-1}\binom{n-2m-1}{i}\tag{10}\\&=2^{n-2m-1}\binom{n-m-1}{m}\end{align*} and the claim follows.
Comment:
In (1) we start with index $i=m$ since $\binom{i}{m}=0$ for $0\leq i<m$. We also set the upper limit of the series to $\infty$ without changing anything since we are adding zeros only. We also use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.
In (2) we shift the index to start from $i=0$.
In (3) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
In (4) we apply the coefficient of operator twice.
In (5) we use the linearity of the coefficient of operator and apply the rule \begin{align*}[z^{p-q}]A(z)=[z^p]z^qA(z)\end{align*}
In (6) we apply the substitution rule of the coefficient of operator with $u:=z^2$\begin{align*}A(z)=\sum_{k=0}^\infty a_k z^k=\sum_{k=0}^\infty z^k [u^k]A(u)\end{align*}
In (7) we use $1-z^2=(1-z)(1+z)$.
In (8) we do a binomial series expansion of $(1+z)^{n-m-1}$ and select the coefficient of $z^{n-2m-1}$. We restrict the upper limit of the sum with $n-2m-1$ since the exponent of $z$ is non-negative.
In (9) we select the coeffcient of $z^{n-2m-1-i}$ and use again $\binom{p}{q}=\binom{p}{p-q}$.
In (10) we apply the binomial identity $\binom{r}{m}\binom{m}{k}=\binom{r}{k}\binom{r-k}{m-k}$ and observe that one factor does not depend on the index $i$.
Note: This answer was inspired by this post from @MarkoRiedel.