Finding the system function of a digital filter
Actually, the denominator is not one. Think about the term \$z^{-1} = \frac{1}{z}\$. Given this we have to re-interpret \$H(z)\$:

$$H(z) = a \left(\frac{1}{z} + z\right) + b = a \left(\frac{1 + z^2}{z}\right) + b\frac{z}{z}$$

Now that all these terms have a common denominator we have found a well formed transfer function:

$$H(z) = \frac{az^2 + bz + a}{z}$$

From here we can find the poles by setting the denominator equal to zero, we see quickly that the pole is at the origin (\$z=0\$). (B.T.W. a denominator of 1 does not mean that there is a pole at the origin, it means that there are no poles!) To find the zeros we set the numerator equal to zero and solve for \$z\$.

$$az^2 +bz +a = 0$$

Using the quadratic equation gives the zeros as:

$$z_{1,2}= \frac{-b \pm \sqrt{b^2 - 4a^2}}{2a}$$