The equation with binomial coefficient $\binom{n-m}{k+m}=\binom{n+m}{k-m}$

Find all positive integers $n,k$ such that $$\binom{n-m}{k+m}=\binom{n+m}{k-m}$$

1) I solved problem if $m=1$. Its here: $k=1; n=3$

2) $$\binom{n-m}{k+m}=\binom{n+m}{k-m}$$ $k=m, n=3m$ is root of this equation.

Does this equation have other roots?

I can't figure out how to get the roots, but I have tried for an approximation. Starting with $$\binom{n-m}{k+m} = \binom{n+m}{k-m}$$ This is $$\frac{(n-m)!}{(k+m)!(n-k-2m)!} = \frac{(n+m)!}{(k-m)!(n-k+2m)!} =$$ $$\frac{(n-m)(n-m-1)...(n-k-2m+1)}{(k+m)!} = \frac{(n+m)(n+m-1)...(n-k+2m+1)}{(k-m)!}$$ Note that the first numerator has $k+m$ products while the second has $k-m$ products. $$Numerator_1(k-m)! = Numerator_2(k+m)!$$ $$Numerator_1 = Numerator_2(k+m)(k+m-1)...(k-m+1)$$ There are $2m$ products in the expression following numerator 2. Thus, for $k$ and $n$ large, $m$ and $k$ approximately satisfy $$k + m = 2m(k-m)$$