Can we get Pauli Exclusion Principle from QFT?

A fermion is a state $|\vec p,\sigma\rangle$ with half-integer spin, i.e., such that $$ J|\vec p,\sigma\rangle=\sigma|\vec p,\sigma\rangle\qquad\text{with}\qquad \sigma\in\mathbb N+\frac12 $$ where $J$ is the angular momentum operator (generator of rotations).

Therefore, upon a rotation by an angle $2\pi$ around an arbitrary axis, $$ U(2\pi)|\vec p,\sigma\rangle=\mathrm e^{i\pi}|\vec p,\sigma\rangle=-|\vec p,\sigma\rangle $$

Finally, if you have a two-fermion system, interchanging them is equivalent to rotating the system by an angle $2\pi$, and therefore $$ |\vec p_1,\vec p_2\rangle=U(2\pi)|\vec p_2,\vec p_1\rangle=-|\vec p_2,\vec p_1\rangle $$

To make this suggestive argument precise, one needs the CPT theorem, which can be proven in an axiomatic framework using just a couple of properties of quantum fields. For more details, see PCT, Spin and Statistics, and All That, by Streater and Wightman.

Spin, Statistics, CPT and All That Jazz by Baez might be of interest too (see the last entry for an explanation of what a rotation in imaginary time has to do with the spin and statistics).

The spin-statistics theorem is a consequence of causality in a relativistic QFT. In order the theory to be causal, the commutator of the fields

$\left[ \Phi (x), \Phi(x^{\prime}) \right ]$

must vanish outside the light-cone, namely, for $(x-x^{\prime})^2>0$.

See the detailed discussion of the subject in Ch. 5 of Weinberg's first volume.