Must all models of ZFC (in a standard formulation) be at least countable?

Must all models of ZFC (in a standard formulation) be at least countable?

Why I think this: there are countably many instances of Replacement, and so, if a model is to satisfy Replacement, it must have at least countably many satisfactions of it.

Does my question only apply to first-order formulations of ZFC, or are there second-order formulations of ZFC that can be finite?

Thanks (Bow)

Re: Must all models of ZFC (in a standard formulation) be at least countable?

yes.

i don't think you even need replacement, the axiom of infinity will suffice.