The following sequence is built of $((A^z)^x -1)/6$ where $z$ increases by $1$ and the conditions for $A$ must be that $(A - 1) \mod 6 = 0$ and for $x$ must be that $x$ is odd and $x > 1$:
$((A^1)^x -1)/6 $ , $((A^2)^x-1)/6$ , $((A^3)^x-1)/6$ , $((A^4)^x-1)/6$ ...
Examples:
$(7^3-1)/6 =57$ , $(49^3-1)/6 =19608$ , $(343^3-1)/6 = 6725601$....
$57, 19608, 6725601$ all have a factor of $19$
$(7^5-1)/6 =2801$ , $(49^5-1)/6 = 47079208$ , $(343^5-1)/6 = 791260261657$....
$2801, 47079208, 791260261657$ all have a factor of $2801$
$(13^3-1)/6 =366$ , $(169^3-1)/6 =1 804468$ , $(2197^3-1)/6 = 1767416562$....
$366, 804468, 1767416562$ all have a factor of $61$
Where as if $x$ is allowed to be $x=2$:
$(7^2-1)/6 = 8$ , $(49^2-1)/6 = 400$ , $(343^2-1)/6 = 19608$ , $(2401^2-1)/6 = 960800$....
$8$ has a factor of $2$
$400$ has factors of $2,5$
$19608$ has factors of $2,3,19,49$
$960800$ has factors $2,3,5,1201$
As you can see there is no consistency in a common factor besides $2$.
$(11^2-1)/6 = 20$ , $(121^2-1)/6 = 2440$ , $(1331^2-1)/6 = 295260$....
$20$ has factors of $2,5$
$2440$ has factors of $2,5,61$
$295260$ has factors of $2,3,5,7, 19, 37$
As you can see there is no consistency in a common factor besides $2$ and $5$ ($5$ for obvious reasons related to the multiplications of 11).
The reason I allow myself to compare $x=2$ with other odd values where $x > 1$ is because $2$ is the only even prime.
From there on if $x$ is a multiplication of $2$ it can be rewritten in a way that $x$ is prime factor. Example $3^6=729$ can be written as $9^3$
The comparison is then reduced to a comparison between prime numbers with different $x$, and the result is that when $x=2$, there is no consistent common factor unlike when $x ≠ 2$
Why is there no consistent common factor when $x=2$?
Notes:
*I don’t take into account the common factors of 2,3 and 5
The sequence that you are studying is the sequence of numbers $$\frac{A^{2z}-1}{6}, z=1,2,3,...$$ Since $$A^{2z}-1=(A^2-1)(1+A^2+A^4+ ... +A^{2z-2})$$ there is a consistent common factor of $$\frac{A^{2}-1}{6}.$$
In your example with $A=7$ this common factor is $8$.
The next $A$ would be $13$ and then the common factor would be $28$ which contains the prime $7$ and so is not in your excluded set of $\{2,3,5\}$.