FIR or IIR filter from poles and zeros
Axel Mancino's answer is correct for causal filters. In general, FIR filters have poles at either $z=0$ or $|z|\rightarrow\infty$, or both.

Take as an example a fourth-order causal FIR filter:

$$H_1(z)=a+bz^{-1}+cz^{-2}+dz^{-3}+ez^{-4}\tag{1}$$

Clearly, $H_1(z)$ has all its poles at $z=0$. An anti-causal FIR filter such as

$$H_2(z)=az^4+bz^3+cz^2+dz+e\tag{2}$$

has all its poles at $|z|\to\infty$. And, finally, a general non-causal FIR filter has poles at $z=0$ as well as at $|z|\rightarrow\infty$:

$$H_3(z)=az^2+bz+c+dz^{-1}+ez^{-2}\tag{3}$$

In sum, a filter with all its poles at $z=0$ or $|z|\to\infty$ (or both) is an FIR filter. Filters with poles in the region $0&lt;|z|&lt;\infty$ must be IIR.