Suppose your population is $\{x_1,...,x_N\}$. The sample without a replacement of a size $m$ is $\{x_{i_1},...,x_{i_m}\}$, where indexes $\{i_1,...,i_m\}$ form a subset of the index set $\{1,...,N\}$. Now there are $m \choose n$ ways of selecting such a subset so your sample has a probability $1/{m \choose n}$ of being drawn. Now
$$Es^2=\sum_{i_1,...,i_m\subset {1,...,N}}s_{i_1,...,i_m}^2\frac{1}{m\choose n},\quad(1)$$
where
$$s^2_{i_1,...,i_m}=\frac{1}{m}\sum_{k=1}^mx_{i_k}^2-\left(\frac{1}{m}\sum_{k=1}^mx_{i_k}\right)^2, \quad (2)$$
Now you must substitute (2) into (1) and show that the expression is equal to
$$\frac{N}{N-1}\left(\frac{1}{N}\sum_{k=1}^Nx_{k}^2-\left(\frac{1}{N}\sum_{k=1}^Nx_{k}\right)^2\right),$$
where I substituted the definition of $\sigma^2$. The task might look daunting, but it is not that complicated. Here are two hints:
$$\sum_{i_1,...,i_k}x_{i_k}^2=\sum_{k=1}^Nx_k^2 {{m-1}\choose {N-1}}$$
$$\sum_{i_1,...,i_k}x_{i_k}x_{i_l}=\sum_{k\neq l}x_kx_l{{m-2}\choose {N-2}}$$
There is also less complicated way of achieving that, but I forgot the details how, I took the survey statistics curse 10 years ago and did not use them afterwards.