For $k<e^{c\sqrt{\log x}}$, you can prove the above using partial summation along with the prime number theorem, but it is provably false for $k\sim x^{u}$ when we assume RH. For a complete solution, take a look at this past answer. Specifically, I prove that $$\sum_{p\leq x}p^{k}=\text{li}\left(x^{k+1}\right)+O\left(x^{k+1}E(x)\right),$$ where $\text{li}(x)=\int_2^x \frac{1}{\log t}dt$ is the logarithmic integral, and where $E(x)$ is any positive increasing function which bounds the error term $\pi(x)-\text{li}(x)$. This gives the asymptotic $$\sum_{p\leq x}p^{k}\sim \frac{x^{k+1}}{(k+1)\log x}$$ uniformely for all $k<e^{c\sqrt{\log x}}$, and if you use the Walfisz bound it can be taken slightly further to $e^{c(\log x)^{3/5}}$ with some doubly logarithmic terms in the exponent.

If you assume the Riemann Hypothesis, this shows that your above expressions is not correct for $k\approx x^u$ where $0<u<1$. On RH we have that $E(x)\ll x^{\frac{1}{2}+\epsilon}$, and so the asymptotic holds for all $k<x^{\frac{1}{2}-\epsilon}$. Then for $k=x^u$ we have $\log x-\log k=(1-u)\log x$, which yields the asymptotic $$(1-u)\frac{x^{k+1}}{(k+1)},$$ which is off by the constant factor $(1-u)$.

There may be a simpler way to prove that your expression is incorrect for $k\sim x^u$ that does not require RH. I think you need to be clever though.

Hope that helps,