From _Altug Alkan_, Jan 21 2017:

Proof: By Dirichlet's theorem, there are infinitely many prime numbers of the form m*n + 1 for some m and any given n. 
So ((m*n + 1)^m)^n is the n-th power refactorable number for infinitely many m. But this proof covers only some prime 
powers that Dirichlet's theorem allows. 

This is the more general proof:

Every number r of the form (rad(m*n + 1)^m)^n is the n-th power refactorable number for m > 0, n > 1 
where rad(n) is the squarefree kernel of n (A007947). Note that r = (rad(m*n + 1)^m)^n is also a 
primitive refactorable number (A235524) for all m > 0, n > 1 according to Zelinsky's definition.

In order to prove this, we must show that d(rad(n + 1)^n) = (n + 1)^omega(n + 1) divides 
rad(n + 1)^n where omega(n) is the number of distinct primes dividing n (A001221). Since 
all exponents in the prime factorization of rad(n + 1)^n are the same and they are n, 
we can focus the maximal exponent in prime factorization of n + 1 in order to show that 
(n + 1)^omega(n + 1) divides rad(n + 1)^n.

Let we define the prime factorization of n + 1 = p^w * q^y * t^z * ... where w is the maximal exponent. 
If (n + 1)^omega(n + 1) divides rad(n + 1)^n, then p^n = p^((p^w * q^y * t^z * ...) - 1) >= (p^w)^omega(n + 1). 
It is only possible with (p^w * q^y * t^z * ...) - 1 >= w * omega(p^w * q^y * t^z * ...). 
Indeed this inequality always works;

(i)   omega(n+1) = 1; p^w >= w + 1.
(ii)  omega(n+1) = 2; p^w * q^y >= 2*w + 1, since p^w * q^y >=  2 * p^w >= 2 * (w + 1) >= 2*w + 1.
(iii) omega(n+1) = 3; p^w * q^y * t^z >= 3*w + 1, since p^w * q^y * t^z >= 2 * (p^w * q^y) >= 2 * (2 * w + 1) >= 3*w + 1.

This situation continues for all values of omega(n+1) thanks to previous inequality and all inequalities work, thus rad(n + 1)^n is refactorable.

If we construct m*n instead of n, we will have an arbitrary parameter m and this makes infinitely many 
n-th power primitive refactorable numbers of the form (rad(m*n + 1)^m)^n for any given value of n > 1.