If $X$ is a random variable with distribution uniform in $[0,1]$, then $E[X^k] = 1/(k+1)$ and $E[X]^k = 1/2^{k}$ so:
$${E\left[X^k\right] \over E[X]^k} \xrightarrow{k\to\infty} \infty$$
What other distributions (of non-negative random variables) have this property?
NOTE: Jensen's inequality implies that for every non-negative random variable $X$: $E\left[X^k\right]\geq E[X]^k$ (is this true?). But it does not imply that the ratio between them goes to infinity.
Let $X$ be a non-negative non (almost surely) constant random variable whose expectation is positive. Then there exists a positive $\varepsilon$ such that $$\mathbb P\left\{\frac X{\mathbb E\left[X\right]} \geqslant 1+\varepsilon \right\}\gt 0 .$$ Otherwise, we would have for any $n$ that $$\mathbb P\left\{\frac X{\mathbb E\left[X\right]} \geqslant 1+\frac 1n \right\}= 0$$ hence $$\mathbb P\left\{\frac X{\mathbb E\left[X\right]} \gt 1 \right\}= 0.$$ As a consequence, defining $Y:= \mathbb E\left[X\right]-X$, the random variable $Y$ is non-negative and its expectation is zero. Therefore, $X$ would be almost surely equal to its expectation, which is excluded.
Now, since $X/\mathbb E\left[X\right]$ is non-negative, it follows that
$$\mathbb E\left[\left( \frac X{\mathbb E\left[X\right]}\right)^n\right] \geqslant \mathbb E\left[\left( \frac X{\mathbb E\left[X\right]}\right)^n\mathbf 1 \left\{\frac X{\mathbb E\left[X\right]} \geqslant 1+\varepsilon \right\}\right] \geqslant \left(1+\varepsilon\right)^n\mathbb P \left\{\frac X{\mathbb E\left[X\right]} \geqslant 1+\varepsilon \right\} \to +\infty. $$