This is very easy to prove using the integral representation of
binomial coefficients. Suppose we are trying to prove that
$$\sum_{k=0}^n {n\choose k} {m+k\choose n}
+ \sum_{k=0}^n {n\choose k} {m+1+k\choose n}
= \sum_{k=0}^{n+1} {n+1\choose k} {m+k\choose n}.$$
Use the two integrals
$${m+k\choose n}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+k}}{z^{n+1}} \; dz$$
and
$${m+k+1\choose n}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+1+k}}{z^{n+1}} \; dz.$$
This yields for the LHS the following sum consisting of two terms:
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z^{n+1}}
\sum_{k=0}^n {n\choose k} (1+z)^k\; dz
\\ + \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{m+1}}{z^{n+1}}
\sum_{k=0}^n {n\choose k} (1+z)^k\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z^{n+1}} (2+z)^n \; dz
+ \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{m+1}}{z^{n+1}} (2+z)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z^{n+1}} (2+z)^{n+1} \; dz.$$
For the RHS we get the integral
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z^{n+1}}
\sum_{k=0}^{n+1} {n+1\choose k} (1+z)^k\; dz
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^m}{z^{n+1}} (2+z)^{n+1} \; dz.$$
The integrals on the LHS and on the RHS are the identical,
QED.
A similar calculation is at this
MSE link.
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
