<s>[INST] identities and binomial coefficients 
I'm having some problems proving this identity. I tried using some formulas I found on the internet so I can turn that $2$ base number into something else but i'm not really sure how to do that. I would be really thankful if someone could help!
 [/INST] We have by inspection that
$$\sum_{k=0}^n {m+k\choose k} 2^{n-k}
= 2^n [z^n] \frac{1}{1-z} \frac{1}{(1-z/2)^{m+1}}.$$
This is
$$2^n \times \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{1-z} \frac{1}{(1-z/2)^{m+1}}
\\ = - 2^n  \times \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{z-1} \frac{2^{m+1}}{(2-z)^{m+1}}
\\ = 2^{n+m+1} (-1)^{m} \times \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{z-1} \frac{1}{(z-2)^{m+1}}.$$
With
$$f(z) = 2^{n+m+1} (-1)^{m} \frac{1}{z^{n+1}}
\frac{1}{z-1} \frac{1}{(z-2)^{m+1}}$$
we will be using the fact that residues sum to zero i.e.
$$\mathrm{Res}_{z=0} f(z)
+ \mathrm{Res}_{z=1} f(z)
+ \mathrm{Res}_{z=2} f(z)
+ \mathrm{Res}_{z=\infty} f(z) = 0.$$
The  residue at  infinity is  zero since  $\lim_{R\to\infty} 2\pi  R /
R^{n+1} / R / R^{m+1} = 0.$
The residue at one is
$$2^{n+m+1} (-1)^{m} \times (-1)^{m+1} = - 2^{n+m+1}.$$
For the residue at two we use the Leibniz rule:
$$\frac{1}{m!} \left( \frac{1}{z^{n+1}}
\frac{1}{z-1} \right)^{(m)}
\\ = \frac{1}{m!} \sum_{k=0}^m {m\choose k}
(-1)^k \frac{(n+k)!}{n!} \frac{1}{z^{n+1+k}}
(-1)^{m-k} \frac{(m-k)!}{(z-1)^{m-k+1}}
\\ =  (-1)^m \sum_{k=0}^m {n+k\choose k}
\frac{1}{z^{n+1+k}} \frac{1}{(z-1)^{m-k+1}}.$$
Restore factor in front and evaluate at $z=2$:
$$2^{n+m+1} (-1)^{m} \times
(-1)^m \sum_{k=0}^m {n+k\choose k}
\frac{1}{2^{n+1+k}}
= \sum_{k=0}^m {n+k\choose k} 2^{m-k}.$$
Summing the residues we have shown that
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=0}^n {m+k\choose k} 2^{n-k}
+ \sum_{k=0}^m {n+k\choose k} 2^{m-k}
- 2^{n+m+1} = 0} $$
which is the claim.
Remark. This is  a special case of an identity  by Gosper which
was         proved        at         the        following         MSE
link (set $x=1/2.$) 
Addendum.  The  initial  step  may be  done  using  an  Iverson
bracket.  We have
$$\sum_{k=0}^n {m+k\choose k} 2^{n-k}
= \sum_{k\ge 0} {m+k\choose k} 2^{n-k} [[0\le k\le n]]
\\ = \sum_{k\ge 0} {m+k\choose k} 2^{n-k}
[z^n] \frac{z^k}{1-z}
= [z^n] \frac{1}{1-z} \sum_{k\ge 0} {m+k\choose k} 2^{n-k} z^k
\\ = 2^n [z^n] \frac{1}{1-z} \sum_{k\ge 0} {m+k\choose k} 2^{-k} z^k
= 2^n [z^n] \frac{1}{1-z}  \frac{1}{(1-z/2)^{m+1}}.$$
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