Since sigma is multiplicative on relatively prime
numbers, sigma(n) can be prime only when n is a power of a prime.
Furthermore, sigma(p^k)=1+p+p^2+..+p^k=[p^(k+1)-1]/[p-1] so if (k+1)
factors, say k+1=ab, then sigma(p^k)=([p^a]^b-1)/(p-1)=([p^a]^b-1)/(p^a-1)*(p^a-1)/(p-1)
is a composite integer. Consequently, k+1 is prime.
In other words, we are looking for four primes p,q,r,s with
(p^q-1)/(p-1)=(r^s-1)/(r-1) with the shared value also a prime. The
example given corresponds to (2^5-1)/(2-1)=(5^3-1)/(5-1).
I searched through q=3 with the following technique: write
M=(r^s-1)/(r-1), so our equation reads (p^3-1)/(p-1)=p^2+p+1=M. Then
an integral value for p requires the discriminant=4M-3 be a perfect
square. I checked for prime s values between 5 and 41 and primes r up
to (so far) a little over 13 million, cacluated M, calculated 4M-3,
then tested whether this was a quadratic residue mod all primes
between 11 and 193. So far I have found no example for the q=3 case
within these constraints.
There was one almost example: if you think of
sigma([-91]^2)=[-91]^2+[-91]+1=8191=2^13-1 and 8191 is prime.
Oh well, probably time to think about higher q values........