Edit: Time to add details, I think. The OP has long since worked it out but hasn't taken the invitation to write up a more complete solution, so I shall, in the interest of having a full answer to the question.
A pivot is a function of the data and the statistic whose distribution doesn't depend on the value of the statistic.
So consider:
(1) what would the distribution of a statistic consisting of the sum of the observations ($T=\sum_i x_i$) be?
A sum of $n$ i.i.d. $\text{gamma}(\alpha,\theta)$ random variables has the $\text{gamma}(n\alpha,\theta)$ distribution (for the shape-rate form of the gamma).
Here $n=6$ and $\alpha=2$, so the sum, $T$ has a $\text{gamma}(12,\theta)$ distribution.
(2) Note that the distribution in (1) does depend on $\theta$ and the form of the statistic doesn't. You need to modify the statistic ($Q=f(T,\theta)$) in such a way that both of those change. (This part is trivial!)
Let $Q=T/\theta$. Then $Q\sim \text{gamma}(12,1)$.
$Q$ satisfies the conditions required for a pivotal quantity.
(3) Once you have a pivotal quantity (i.e. $Q$), write down an interval for the pivotal quantity (in the form of a pair of inequalities, $a< Q< b$) with the given coverage. Since the distribution doesn't depend on the parameter, this interval is always the same (at a given sample size) no matter what the value of $\theta$.
One such interval is $(a,b)$, where $P(a<Q<b)=0.95$, when $a$ is the 0.025 point of the $\text{gamma}(12,1)$ distribution and $b$ is the 0.975 point.
(4) Now write the interval involving the pivotal quantity back in terms of the data and $\theta$. Back out an interval for the parameter, for which the corresponding probability statement must still hold (keeping in mind that the random quantity is not $\theta$ but the interval).
$P(a<T/\theta<b)=0.95$ implies $P(1/b < \theta/T < 1/a)=0.95$, so $P(T/b < \theta < T/a)=0.95$. Therefore $(T/b,T/a)$ is a 95% interval for $\theta$.
Our observed total, $t = 4.91$. The 0.025 point of a gamma(12,1) is 6.2006 and the 0.975 point is 19.682. Hence a 95% interval for $\theta$ is (4.91/19.682,4.91/6.200)
= $(0.249, 0.792)$.
Best Answer
It appears that you are confusing yourself by bringing in the pivotal quantity $Z$ that comes from a completely different type of distribution. That quantity does not arise in this problem, since you have only one observation, and the parameters in that pivotal quantity are not defined in this problem.
In general terms, a pivotal quantity is just a function of the observable data and parameters that has a distribution that does not depend on the parameters. So, in this question, once you have shown that $Y$ has a distribution that does not depend on $\theta$, you have shown that $Y$ is a pivotal quantity ---i.e., there is nothing left for you to do. In this case, the "hint" you were given is effectively giving you the pivotal quantity, and all you needed to do was show that its distribution does not depend on $\theta$.
Confirming the pivotal quantity: I am getting the same answer as you for the distribution, but it is a good idea to specify the support of the distribution. For all $0 \leqslant y \leqslant 1$ we have:
$$\begin{equation} \begin{aligned} F_Y(y) \equiv \mathbb{P}(Y \leqslant y) &= \mathbb{P}(\tfrac{\theta-X}{\theta} \leqslant y) \\[6pt] &= \mathbb{P}(X \geqslant (1-y) \theta) \\[6pt] &= \int \limits_{(1-y)\theta}^\theta \frac{2 (\theta-x)}{\theta^2} dx \\[6pt] &= \Bigg[ \frac{x (2 \theta - x)}{\theta^2} \Bigg]_{x=(1-y)\theta}^{x=\theta} \\[6pt] &= 1 - 2 (1-y) + (1-y)^2 \\[6pt] &= 1 -2 + 2y + 1-2y+y^2 \\[6pt] &= y^2. \\[6pt] \end{aligned} \end{equation}$$
This gives the corresponding density function:
$$f_Y(y) = 2y \quad \quad \quad \text{for } 0 \leqslant y \leqslant 1.$$
Since $f_Y$ does not depend on the parameter $\theta$, the function $Y$ is a pivotal quantity in this problem. Note that this quantity has no particular relationship with $Z$, which is the pivotal quantity from an entirely different problem.
Using the pivotal quantity: It might be useful for you to understand that pivotal quantities are used to form confidence intervals. This is done by forming a probability statement on the pivotal quantity and then "inverting" this statement to make it a statement about the location of the parameter of interest. In the present case, for any value $0 < \alpha < 1$ we can form the probability statement:
$$\begin{equation} \begin{aligned} 1-\alpha &= \mathbb{P}(0 \leqslant Y \leqslant \sqrt{1-\alpha}) \\[6pt] &= \mathbb{P}(0 \leqslant 1-\tfrac{X}{\theta} \leqslant \sqrt{1-\alpha}) \\[6pt] &= \mathbb{P}(1-\sqrt{1-\alpha} \leqslant \tfrac{X}{\theta} \leqslant 1 ) \\[6pt] &= \mathbb{P} \Big( X \leqslant \theta \leqslant \frac{X}{1-\sqrt{1-\alpha}} \Big). \\[6pt] \end{aligned} \end{equation}$$
Substituting the observed value $x$ gives the following $1-\alpha$ level confidence interval for $\theta$:
$$\text{CI}_\theta(1-\alpha) = \Big[ x, \frac{x}{1-\sqrt{1-\alpha}} \Big].$$