Denote the statement "there exists a standard model" as $\mathsf{SM}$. I know that $\operatorname{Con}(\mathsf{ZFC})$ implies $\operatorname{Con}(\mathsf{ZFC} + \neg \mathsf{SM})$ because given a model $M$ of ZFC, if $M$ doesn't already satisfy $\neg \mathsf{SM}$, we can take the minimal submodel of $M$. I also know that in a model of $\mathsf{ZFC} + \neg \mathsf{SM}$, there will still exist (non-standard) models of $\mathsf{ZFC}$ assuming $\operatorname{Con}(\mathsf{ZFC})$. So my question is:
- Does some of those non-standard models necessarily satisfy $\mathsf{SM}$, or
- Is it possible that all of those non-standard models also satisfy $\neg \mathsf{SM}$, or
- Is there some reason it's difficult to say either way?
The connection of the question in the title to the question in the body is unclear to me, but the answer to the question in the title is straightforwardly "no" as even $\sf ZFC + Con(ZFC)$ is strictly higher consistency strength than $\sf ZFC$, so certainly $\sf ZFC + SM$ is.
As for the question in the body, say we're in the minimal standard model of $\sf ZFC$. Even though this model doesn't satisfy $\sf SM$, it's standard, so still satisfies all true statements of arithmetic, so satisfies $\sf Con(ZFC + SM)$ and $\sf Con(ZFC + \lnot SM)$, and hence some of the (necessarily) nonstandard models of $\sf ZFC$ in the minimal standard model will satisfy $\sf SM$ and others will not.
However, if we're just in some nonstandard model $M$ of $\sf ZFC + \lnot SM$, all bets are off. You didn't specify that $M$ satisfies $\sf Con(ZFC)$, so there could be no models of $\sf ZFC$ at all, according to $M$. Or, $M$ could be be a model of $\sf ZFC$ that thinks $\sf ZFC$ is consistent, but thinks $\sf ZFC + SM$ is $\mbox{inconsistent}^*$ so then according to $M$, there are models of $\sf ZFC$ but none of them satisfy $\sf SM$.
$^*$ It occurs to me the entire point of the question may be to ask if this scenario is possible (and the title question makes more sense in this light), so worth a few more words on why such models exist. It's an immediate consequence of the incompleteness theorem, applied to $\sf ZFC + Con(ZFC)$, which implies there are models of $\sf ZFC + Con(ZFC) + \lnot Con( ZFC + Con(ZFC) )$. Such a model will satisfy that there are models of $\sf ZFC$, but also that none of these models think there are any models of $\sf ZFC$, let alone standard models. Note that such a model necessarily satisfies $\sf \lnot SM$, since any standard model of $\sf ZFC$ satisfies $\sf ZFC + Con(ZFC).$