The given equation is equivalent to $$\begin{align*}\frac{n!}{(k-1)!(n-k+1)!} &=2\cdot\frac{n!}{k!(n-k)!} + \frac{n!}{(k+1)!(n-k-1)!}\end{align*}$$

Multiplying both sides by $\frac{(k+1)!(n-k+1)!}{n!}$ yields $$k(k+1)=2(k+1)(n-k+1)+(n-k)(n-k+1)$$

which implies $$(2n+3)^2 - 2(2k+1)^2 = 1$$

This is a standard Pell's equation. Its fundamental solution is $$(x, y) = (2n+3, 2k+1) = (1,1)$$

and all the solutions $(x_i, y_i)$ are given by $$x_i + y_i \sqrt{2} =(1+\sqrt{2})^{2i+1}$$

where $x_i = 2n_i+3$ and $y_i=2k_i+1$. Note that $x_i$ and $y_i$ are always odd. Therefore corresponding $n$ and $k$ are always integers.