Combinatorial proof ($n > 0, m > 1$):
$$\sum_{k=0}^n {{n}\choose{k}}^2 \ m^k = \sum_{j=0}^n {{n}\choose{j}} \ {{2n - j}\choose{n}} \ (m-1)^j$$
I came up with a story that work only for one side of the equation and doesn't for the another.
Starting from the LHS:
W are enumerating ordered trios of elements $(k, L_1, L_2)$ where $k$ is an integer $\in [0,n]$ and $L_1$ is a group of $k$ elements $\in [0,n]$ and $L_2$ is a group of $k$ elements $\in [0,n]$ each holding a value $\in [1,m]$. In that scenario, the choices of elements making $L_1$ and $L_2$ are independent of each other.
That type of thinking doesn't seem to work for the RHS. There I am choosing from $2n-j$ and that looks like a set of combined $2$ sets of $n$ elements without $j$ elements that I have chosen previously. That seems to indicate that both choises should be dependent on each other.
Also I have no idea why on the RHS there would be $(m-1)^j$ if on the LHS I have $m^k$. All things aside, those should be the same since we are talking about the choice of values from set $m$ which is not dependent on anything in a summation.
Here is a partial answer: handling the powers algebraically and the coefficients by a combinatorial proof.
Lemma: for $0\le j\le n$ we have $$\sum_{k=j}^n \binom nk^2\binom kj=\binom nj\binom{2n-j}n\ .$$
Proof. Choose a set $A$ of size $j$ from $\{1,\ldots,n\}$ and a set $B$ of size $n$, disjoint from $A$, from $\{1,\ldots,2n\}-A$. Doing it the obvious way gives the right-hand side. Alternatively, choose $B$ by picking $k$ elements from $\{1,\ldots,n\}$ which will not be in $B$, and $k$ from $\{n+1,\ldots,2n\}$ which will. Then from the former choose $j$ elements for $A$. This requires $j\le k\le n$ and gives the left-hand side.
Then for the algebra: to simplify notation I write $m+1$ instead of your $m$. Then $$\eqalign{ \sum_{k=0}^n\binom nk^2(m+1)^k &=\sum_{k=0}^n\binom nk^2\sum_{j=0}^k \binom kj m^j\cr &=\sum_{j=0}^n\Bigl(\sum_{k=j}^n \binom nk^2\binom kj\Bigr)m^j\cr &=\sum_{j=0}^n \binom nj\binom{2n-j}n m^j\ .\cr}$$