Some steps I have done: $$P(Z\le i) = P(X \le i) P(Y \le i) = (1-(1-p)^m)(1-(1-p)^m)$$ $$P(Z =i) = P(Z \le i) - P(Z \le i - 1)$$ I am not able to write $P(Z =i)$ in the form of $q(1-q)^{k-1}$.
Since $$\Pr[Z \le i] = (1 - (1-p)^i)^2,$$ then $$\Pr[Z = i] = (1 - (1-p)^i)^2 - (1 - (1-p)^{i-1})^2 = (2-(2-p)(1-p)^{i-1})(1-p)^{i-1} p,$$ where $i \in \{1, 2, 3, \ldots\}.$ This cannot be written as a geometric distribution because $Z$ is not geometrically distributed.
To prove that the PMF for $Z$ is not geometric, note that for any geometric distribution, we must have $$\frac{\Pr[Z = i+1]}{\Pr[Z = i]} = 1-p,$$ which is constant with respect to $i$. But in this case, we can easily see that for $p = 1/2$, the PMF of $Z$ is $$\Pr[Z = i \mid p = 1/2] = \frac{2^{i+1} - 3}{4^i},$$ hence $$\frac{\Pr[Z = 3 \mid p = 1/2]}{\Pr[Z = 2 \mid p = 1/2]} = \frac{13/64}{5/16} = \frac{13}{20},$$ but $$\frac{\Pr[Z = 2 \mid p = 1/2]}{Pr[Z = 1 \mid p = 1/2]} = \frac{5/16}{1/4} = \frac{5}{4}.$$ If $Z$ were geometric, these ratios would be equal.
Does $Z$ also geometric distributed?
No, because if you look at this link, you see that the expected value of the maximum does not have the form you'd expect it to have, if it were geometrically distributed.
The CDF is the maximum is the product of the individual CDFs (see this link for a little bit of discussion), which doesn't turn out to be geometric.
You have computed $P(Z \leq i) = (1-(1-p)^i)^2$. To show $Z$ is not geometrically distributed, it suffices to show that there exists no $q \in (0,1]$ such that $(1-(1-p)^i)^2 = 1-(1-q)^i$ for each positive integer $i$, since the right-hand side is the CDF of a geometric random variable with generic parameter $q$.
Plug in $i=1$ to deduce $p^2=q$. Thus $(1-(1-p)^i)^2 = 1-(1-p^2)^i$. If $i=2$, this reads $(1-(1-p)^2)^2 - (1-(1-p^2)^2)=0$, which only has positive solution $p=1$. We can check that if $p=1$, $Z \sim \text{Geometric}(1)$, but of course this edge case is uninteresting.
