Check if a filter is IIR or FIR
In the general case you have

$$H(z)=\frac{P(z)}{Q(z)}$$

where $P(z)$ and $Q(z)$ are polynomials in $z$. If - as is the case in your example - $Q(z)$ just has one single term, $H(z)$ is definitely FIR, because you can simply divide each term of $P(z)$ by that respective power of $z$, and the number of terms of $H(z)$ equals the number of terms of $P(z)$.

If $Q(z)$ is a polynomial with finite zeros away from $z=0$, then $H(z)$ is generally IIR, except if all zeros of $Q(z)$ are cancelled by zeros of $P(z)$. One example of such a case is a recursive moving average (where $N$ is the window length):

$$H(z)=\frac{1}{N}\frac{1-z^{-N}}{1-z^{-1}}\tag{1}$$

The zero of the denominator is cancelled by one of the zeros of the numerator, so the transfer function $(1)$ can equialenty be written as 

$$H(z)=\frac{1}{N}(1+z^{-1}+\ldots+z^{-N+1})\tag{2}$$

which is obviously FIR.