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