Are there positive integers $a,b,c$ such that both $2a(b^c+1),2a(b^c-1)$ are perfect powers? Are there positive intgers $a,b,c$ such that both $2a(b^c+1),2a(b^c-1)$ are perfect powers?

My research
We have $2a(b^c+1)-2a(b^c-1)=4a$, so I looked at perfect powers which have a multiple of $4$ as a difference. I found this question which talks about differences of powers, and I found:
$5^3-11^2=4$
$47^2-13^3=12$
$312^2-46^3=8$
$2^{17}-362^2=28$
and a few more, but none of them seemed to be in our form.
I also found On the Diophantine equation $m^2 - p^k = 4z$, where $z \in \mathbb{N}$ and $p$ is a prime satisfying $p \equiv k \equiv 1 \pmod 4$ which is kind of similar to what are talking about. They say we don't have a solution to their problem in the special case due to Catalan's conjecture, so I guess we also don't have one for our problem, but couldn't prove it.
 A: If $b$ is even  then $b^c+1$ and $b^c-1$ are coprime, and so we have
$$b^c+1=\prod p^{u_p}\qquad\text{ and }\qquad b^c-1=\prod q^{v_q},$$
where the $p$ and $q$ are all distinct prime numbers, and the $u_p$ and $v_q$ are distinct positive integers. To find a positive integer $a$ such that $2a(b^c+1)$ and $2a(b^c-1)$ are both perfect powers, it suffices to find nonnegative integers $s_p$ and $t_q$ such that
$$u_p+s_p\equiv t_q\equiv0\pmod{m}\qquad\text{ and }\qquad v_q+t_q\equiv s_p\equiv0\pmod{n},$$
for some integers $m,n>1$, as then $a:=2^{mn-1}\prod p^{s_p}\prod q^{t_q}$ will make the two products $2a(b^c+1)$ and $2a(b^c-1)$ perfect $m$-th and $n$-th powers, respectively. By the Chinese remainder theorem, this can always be done by choosing $m$ coprime to the $v_q$, and $n$ coprime to the $u_p$, and $m$ and $n$ coprime to eachother.
For a concrete example, take $b=6$ and $c=3$ so that
$$b^c+1=217=7\times31\qquad\text{ and }\qquad b^c-1=215=5\times43.$$
Then $u_7=u_{31}=1$ and $v_5=v_{43}=1$. We may simply take $m=2$ and $n=3$, for which the smallest $s_p$ and $t_q$ are
$$s_7=3,\quad s_{31}=3,\quad t_5=2,\quad t_{43}=2,$$
corresponding to
$$a=2^{2\times3-1}\times(7^3\times31^3)\times(5^2\times43^2)=15114928589600,$$
with
\begin{eqnarray*}
2a(b^c+1)&=&80993080^2,\\
2a(b^c-1)&=&186620^3.
\end{eqnarray*}
Easy exercise: Can you find an appropriate value for $a$ given $(b,c)=(2,21)$?
Harder exercise: Can you extend this construction to odd $b$?
