Can ZF set theory prove that every infinite set has infinitely many partial orders?

Sure. ZF can show an infinite set has an infinite power set. So if $X$ is an infinite set, then for any $Y\subseteq X,$ let $\le_Y$ be the partial order on $X$ defined as $a\le_Y b$ if and only if $a=b$ or $a\in Y$ and $b\in X\setminus Y.$

EDIT

Come to think of it, why not simply take, for any $x\in X,$ $\le_x$ the partial order where $x$ is the minimum and everything else is above $x$ and maximal ($\le_{\{x\}}$ in the notation of my original example). That's simpler and doesn't rely on quoting the result that the power set is infinite, although the injection $x\mapsto \{x\}$ is proof the power set is infinite, so it's not much of a leap.

(But on the other hand, the initial answer, combined with the ZF-provable fact that the power set of an infinite set contains finite sets of all sizes, gives the stronger result that there are infinitely many partial orders up to isomorphism.)