I found no other solution with p < 10^10.
There is a simple heuristic argument to show that our product
cannot be a square : write p+q as a*x^2 and q+r as b*y^2 with a,
b squarefree. If (p+q)*(q+r) has to be a square, we must have a
= b. From y > x it follows : r - p = a*(y^2 - x^2) > 2*a*x.
But, we have x^2 = (p+q)/a > 2p/a and thus, r - p >
2a*Sqrt(2p/a) >= 2*Sqrt(2p), an equality I never saw to hold for any
prime (and which would contradict Legendre's conjecture).
A similar argument for third powers is not possible since there
is the exception 5, 7, 11.
Here is a list of all p < 10^10 for which (p+q)*(q+r) is a
powerful number :5, 7, 43, 139, 283, 1069, 2417, 2591, 3593, 114229,
133831, 151247, 1404479, 12880121, 12967153, 24329119, 43804793.
Since my last entry is far beyond 10^10, it is possible that
this sequence is finite.