Alternative combinatorial proof for $\sum\limits_{r=0}^n\binom{n}{r}\binom{m+r}{n}=\sum\limits_{r=0}^n\binom{n}{r}\binom{m}{r}2^r$

An approach using common binomial identities: $$ \begin{align} \sum_{r=0}^n\binom{n}{r}\binom{m+r}{n} &=\sum_{r=0}^n\binom{n}{r}\sum_{k=0}^n\binom{m}{k}\binom{r}{n-k}\tag{1}\\ &=\sum_{k=0}^n\binom{m}{k}\sum_{r=0}^n\binom{n}{r}\binom{r}{n-k}\tag{2}\\ &=\sum_{k=0}^n\binom{m}{k}\sum_{r=0}^n\binom{n}{n-k}\binom{k}{r-n+k}\tag{3}\\ &=\sum_{k=0}^n\binom{m}{k}\binom{n}{k}2^k\tag{4} \end{align} $$ Explanation:
$(1)$: $\sum\limits_{k=0}^n\binom{m}{k}\binom{r}{n-k}=\binom{m+r}{n}$
$(2)$: change order of summation
$(3)$: $\binom{n}{m}\binom{m}{k}=\binom{n}{k}\binom{n-k}{m-k}$
$(4)$: $\sum\limits_{k=0}^n\binom{n}{k}=2^n$

In fact both numbers are equal to $D_{m,n}$, which is the so-called Delannoy number of an $(m \times n)$-lattice. It is defined to be the number of paths from $(0,0)$ to $(m,n)$ using the steps $(0,1), (1,0)$ and $(1,1)$.

LHS: Classify the paths by the number of $(1,1)$-steps. If there are $r$ diagonal steps, then we need $m+n-r$ steps in total (of which $r$ are diagonal steps) and we have $m+n-2r$ non-diagonal steps (of which $n-r$ are $(0,1)$-steps). That should give you a clue how to go about showing that $LHS=D_{m,n}$.

RHS: Hopefully there is some clever way of counting the number of paths in a different way, showing that $RHS=D_{m,n}$.

Alternatively, maybe one (algebraic) approach could be to show that both expressions satisfy the recurrence relation $$D_{m,n}=D_{m,n-1}+D_{m-1,n}+D_{m-1,n-1}$$

with intial condition $D_{0,0}=1$.

(This follows trivially from the definition of $D_{m,n}$. Just condition on the type of the last step.)