One small solution is p=28, q=73, r=4.
It's easy to generate an infinite number of solutions from Mersenne factors, although they grow rather quickly.
When Robert Silverman left off, we have a solution for any choices of m. k, q, and h such that
m=k+2^{h} and (kq+1)=2^{mh}
If we let z=mh, we have a solution for any values of z, k, q, and h with
k=z2^{h}+h and kq=2^{z}1
Eliminating k, we see that we have a solution for any choice of z, q, and h with
q=(2^{z}1)/(z2^{h}+h)
z cannot be a prime number because the denominator is smaller than z and all factors of 2^{p}1 are of the form 2ap+1, and therefor larger than p.
So let z=xy and pick the denominator to be a factor of 2^{x}1.
So to generate a solution:
1. Pick a value for h.
2. Pick x to be a factor of 2^{h}h+1
3. Pick the denominator to be a factor of 2^{x}1
4. Solve for y from the denominator choice
5. z=x*y
6. Work backwards to p, q, and r.
For example
1. h=2
2. 2^{h}h+1=3, so pick x=3
3. 2^{x}1=7, so pick the denominator=7
4. y = 3
5. z = xy = 9
6. q = (2^{z}1)/denominator = 73
7. m = z+h = 11
8. k = denominator = 7
9. r = 2^{h}= 4
10. p = kr = 28
