Solve the Diophantine equation $ (3^{n}-1)(3^{m}-1)=x^{r} $

Question

Is there a solution other than solution

\begin{equation*} (3^{2}-1)(3^{2}-1)=2^{6}=4^{3} \end{equation*}

of the Diophantine equation \begin{equation*} (3^{n}-1)(3^{m}-1)=x^{r} \end{equation*} for positive integers $ n, m, x, r $ such as $ n, m, x \geq 2 $ and $ r \geq 3 $ ?

Answer 1

these ought to be relatively rare, by Zsigmondy. However, in that theorem, we are not told the exponent on the new prime. So, for example, $3^5 - 1 = 2 \cdot 11^2.$ The new prime, $11,$ is squared, and we find $$ (3^2 - 1)(3^5 - 1) = 44^2 $$

Usually, a new prime is has exponent $1.$

Made a list, my program was able to completely factor and show the new prime(s) until $n$ became a prime over 35... I could do those in gp-pari

Sat Jul  2 17:56:42 PDT 2022
1    2 =  2
2    4 = 2^2
3    13 =  13
4    5 =  5
5    121 = 11^2
6    7 =  7
7    1093 =  1093
8    41 =  41
9    757 =  757
10    61 =  61
11    88573 = 23  3851
12    73 =  73
13    797161 =  797161
14    547 =  547
15    4561 =  4561
16    3281 = 17  193
17    64570081 = 1871  34511
18    703 = 19  37
19    581130733 = 1597  363889
20    1181 =  1181
21    368089 =  368089
22    44287 = 67  661
23    47071589413 = 47  1001523179
24    6481 =  6481
25    3501192601 = 8951  391151
26    398581 =  398581
27    387440173 = 109 433  8209
28    478297 = 29  16493
29    34315188682441 = 59 28537  20381027
30    8401 = 31  271
31    308836698141973 = 683 102673  4404047
32    21523361 =  21523361
33    2413941289 =  2413941289
34    32285041 = 103 307  1021
35    189150889201 = 71  2664097031
36    530713 =  530713
37    225141952945498681 =  cdot mbox{BIG} =  13097927  17189128703  
38    290565367 = 2851  101917
39    15040635637 = 313 6553  7333
40    42521761 =  42521761
41    18236498188585393201 = 83  cdot mbox{BIG}  = 83 2526913 86950696619 
42    97567 = 43  2269
43    164128483697268538813 = 431  cdot mbox{BIG} = 431  380808546861411923
44    3138105961 = 5501  570461
45    271983020401 = 181 1621  927001

Answer 2

Let $ d= \gcd(3^{n}-1; 3^{m}-1) $, so $ d = 3^{\gcd(n;m)}-1 $

solve the diophantine equation \begin{equation} (3^{n}-1)(3^{m}-1)=x^{2} \end{equation} equivalent solving the system a of two Diophantine equations as follows: \begin{equation} 3^{n}-1=d \: \alpha^{2} \\ 3^{m}-1= d \: \beta^{2} \end{equation}

with $ \gcd(\alpha; \beta)= 1 $ and $ x= d \alpha \beta $.