Other people did give, in a way or another, the right practical answer, i.e. you should always consider the unavoidable internal resistance of the battery and/or that of the wire.
From a purely theoretical POV, instead, when you short an ideal voltage source you get a singular, degenerate circuit, with quantities like the current into the short going to infinity. It's the dual case of an ideal current source with no load attached, i.e. with the terminal left open: you get an infinite voltage across them.
This kind of things happen only because ideal sources are just mathematical abstractions, which may give you infinite quantities when used carelessly.
Keep in mind that ideal circuit elements are useful, and simplified, mathematical models of real things. They are not physical objects, so they don't need to follow the law of physics. In particular, ideal sources may source infinite power and this means infinite energy in any finite interval of time: of course this is physically absurd.
In particular, if you fail to model a physical situation correctly, i.e. you oversimplify the model, you can incur in such mathematical absurdities.
If, on the other hand, you are just putting mathematically ideal elements together, there is no need for them to give you something coherent. You are simply writing equations in a graphical ways. Nothing stops you from writing a system of equations with two incompatible equations. The system will have no solution (or will have absurd, non-finite, solutions).
For your specific case, you are just writing two equations and putting them in a system:
1) the equation of the ideal source, saying that the voltage across its terminal (call them A and B) is given and known and different from 0 (call it Vs);
2) the equation of an ideal wire, i.e. a resistance with R=0, which must follow Ohm's law: i.e. I=Vab/R
There is no solution for such a system. Or you may consider I=∞ a "solution", but that is no real number and for any practical purpose is nonsense.