See that $\dfrac{k^{m+1}-1}{(m+1)k^m}= \dfrac{ \int_1^{k} x^m\,dx.}{k^m} > \frac{1^m+2^m+3^m+\ldots +(k-1)^m}{k^m} \ge \dfrac{ \int_0^{k-1} x^m\,dx.}{k^m}=\dfrac{(k-1)(1-\dfrac{1}{k})^m}{m+1}$. The max value of $k$ upon which the Left thinggy is less than $2$ is $2m+2$ (easy to check, do some division ). Hence, it suffices to prove that $k=2m+2$ works. Note tho that by Bernoulli's ($\dfrac{-1}{k}>-1$) , it is $\dfrac{(k-1)(1-\dfrac{1}{k})^m}{m+1}=\dfrac{(2m+1)(1-\dfrac{1}{2m+2})^m}{m+1} \ge \dfrac{(2m+1)(1-\dfrac{m}{2m+2})}{m+1}$, which turns out to be greater than $1$. Hence $k=2m+2$ works.

Let $S_k=1^m+2^m+...+(k-1)^m-k^m$.It is enough to show that there exists positive integer $k$ such that $S_k\in [0,k^m)$. It is known that $1^m+2^m+...+n^m$ is a $m+1$ degree polynomial with positive leading coefficient .So letting $k\to\infty$ there is $N$ such that for all $k\ge N$ $S_k$ is positive.Let $r$ be the biggest integer such that $S_r$ is negative.Note that $S_2$ is negative.So such $r$ exists. Now $S_{r+1}-S_r=2r^m-(r+1)^m<r^m$.As $S_{r+1}$ is positive and from the last inequality it is obvious that $S_{r+1}\in[0,r^m)$.$\blacksquare$