Obviously the question in the title alone doesn't make sense so I'll develop on the context and then I'll ask my question :
Studying $AD$ (axiom of determinacy) I had to prove that $AD$ and $AC$ are incompatible (mod $ZF$). So to prove this I show that under $AC$, there are some undetermined games, and in order to prove this I use the fact that the Fréchet filter can be extended to an ultrafilter (necessarily non principal). As a remark, I thought "This shows that $AD$ is incompatible, not only with $AC$, but also with *strictly* weaker choice principles, such as the $BPI$ or the statement 'There exists a non-principal ultrafilter over $\omega$' ". Thinking this I wondered how strong that last statement (let $\Omega$ denote said last statement) was, and whether it could imply $BPI$. The answer seems to be "Obviously, no", as $\Omega$ is only about a specific set ($\omega$), whereas the $BPI$ is much more general. But then I wondered whether there was any "general choice principle", strong enough to prove $\Omega$, but not strong enough to prove the $BPI$. I then looked for a way to make "general choice principle" precise in order to look for an answer. So here are my questions :

Is there a satisfactory way to make the notion of "general choice principle" precise, i.e. characterize certain sentences such that $AC$, $BPI$, etc. fall under this scope, but $\Omega$ doesn't ?

If there is, let $\phi$ be such a general choice principle. Can we have $ZF + \phi \vdash \Omega$, but $ZF + \phi$ doesn't prove $BPI$ ?

I first thought of defining it as "$ZF^{-}+ \phi$ ($ZF^{-}$ being $ZF$ minus the axiom of infinity) does not prove the axiom of infinity", but 1.it was only for this particular example, 2. it didn't work, as "$Inf \implies \Omega$" would fall under this notion, but clearly wouldn't be satisfcatory, so a solution would have to be more clever than that.