Let $P(x)=x^k+\dfrac{1}{x^k}-x-1$. First, we show that $P\left(1+\dfrac{1}{k}\right)>0$.
For $k=2$, we have $P\left(1+\dfrac{1}{2}\right)=\dfrac{7}{36}>0$.
For $k\ge 3$, the Binomial Theorem implies that
$\left(1+\dfrac{1}{k}\right)^k>1+k\left(\dfrac{1}{k}\right)+\dfrac{k(k-1)}{2}\cdot\dfrac{1}{k^2}=2+\dfrac{k-1}{2k}\ge 2+\dfrac{1}{k}$
Thus, $P\left(1+\dfrac{1}{k}\right)>\left(1+\dfrac{1}{k}\right)^k-\left(1+\dfrac{1}{k}\right)-1>0$
Next, we let $Q(x)=x^kP(x)=x^{2k}-x^{k+1}-x^k+1$. Then it is easy to verify that $Q(x)=(x-1)f(x)$, where
$f(x)=x^{2k-1}+x^{2k-2}+\cdots+x^{k+2}+x^{k+1}-x^{k-1}-x^{k-2}-\cdots-x-1$
Since $P\left(1+\dfrac{1}{k}\right)>0$, we deduce that $Q\left(1+\dfrac{1}{k}\right)>0$ and $f\left(1+\dfrac{1}{k}\right)>0$. But $f(1)=k-1-k=-1<0$, thus the Intermediate Value Theorem implies that there exists $m\in\left(1,\,1+\dfrac{1}{k}\right)$ such that $f(m)=0$. It then follows that $P(m)=0$.