If we define $$f(n):=\frac{\frac{\ln(\sigma_0(n^2-1))}{\ln(n^2-1)}\cdot \ln(ln(x^2-1))}{1.5379 \ln(2)}$$ the above inequality transofrms for an integer $n>1$ to $f(n)<1$. In this sense , the bound is quite sharp since there are relatively large examples exceeding value $0.96$ (I do not know whether there is an infinite family with such a value). The cases I found so far are :
71 0.96550569958544977763836549443179567483
181 0.96430719958309823057999129716333892573
701 0.96833757276827017005723426787533928295
881 0.96941049409320738878398020010442332797
1079 0.97056273964410150770515566495752407562
1189 0.97359958249063265372340150061912548500
3079 0.97145701755800634337156832705326074155
4159 0.98789091805010901983501788042129751432
5851 0.98073284674268129366055688857224013339
11969 0.97329786954312637751860416545229031760
23561 0.96059247135619823274667375530672339754
23869 0.96878708863622687486668378057064844319
24751 0.98288976062213809784757908954476329291
43471 0.98650935298602568177274182459864177715
82081 0.96860919540791328857766234269880534432
94249 0.96949158921503843568531671139381359153
104651 0.96350181878778144071934670069527232375
313039 0.96174492389186480169209535858686197197
32185999 0.96567077627146669252902273427154090478
The last example is particular large. In this sense, the above bound can be considered to be quite good and in this sense , it can probably not be improved much. But this is a logarithmic bound , so there is still room for better bounds concerning the actual value , but I have no idea how we can get such bounds.