The physically measurable quantity in QM is the probability density. This gives you $\left|\psi\left(x,y\right)\right|^2 = \left|\psi\left(y,x\right)\right|^2$ (since when we say two particles are indistiguishable, what we actually mean is that we cannot make any measurements to distinguish them). This would imply that $\psi\left(x,y\right) = \pm\psi\left(y,x\right)$, which tells you that the wavefunction must be odd or even.
As pointed out by eranreches, the reasoning above is flawed, since the wavefunction with particles interchanged could be multiplied by a complex phase factor, in general.
A better way of looking at the above problem would be to define a particle exchange operator $P$ such that $P\psi\left(x,y\right) = \psi\left(y,x\right)$. If the system does not change under the action of this operator, it means that the Hamiltonian is invariant under the action of this operator, which implies $\left[H,P\right] = 0$. This implies that the energy eigenstates are also eigenstates of $P$. We can also see that by definition, $P^2 = I$, where $I$ is the identity operator. Let $p$ be the eigenvalue for $P$ of some energy eigenstate $\psi\left(x,y\right)$. We can then see that $P^2\psi\left(x,y\right) = p^2\psi\left(x,y\right) = I\psi\left(x,y\right) = \psi\left(x,y\right)$. This implies $p^2 = 1$, which implies $p = \pm 1$. Thus we get $\psi\left(y,x\right) = P\psi\left(x,y\right) = \pm\psi\left(x,y\right)$.
This is not the complete story, however. If you want to delve further into why particles with half-integer spins have antisymmetric wavefunctions, and why particles with integer spin have symmetric wavefunctions, you would have to look at the Spin statistics theorem, as pointed out by ACuriousMind.