For integers $n\geq 1$ in this post we denote the square-free kernel as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,$$ that is the product of distinct primes dividing an integer $n>1$ with the definition $\operatorname{rad}(1)=1$ (the Wikipedia's article dedicated to this multiplicative function is Radical of an integer).
Question. I wondered about if there are many much or few integers $n\geq 1$ for which $$n^{\operatorname{rad}(n)}+\operatorname{rad}(n)+1\tag{1}$$ has no repeated prime factors. What work can be done, calculations, heuristics or reasonings, about if the sequence $(1)$ does contain infinitely many terms without repeated prime factors (that is integers similar than next Examples)? Many thanks.
Examples.
1) For $n=1$ the integer expressed as $(1)$ has no repeated prime factors since $1^{\operatorname{rad}(1)}+\operatorname{rad}(1)+1=3$ that is a prime number.
2) For $n=4$ one has also that the corresponding integer of the form $(1)$ is a square-free integer, that is has no repeated prime factors since $4^{\operatorname{rad}(4)}+\operatorname{rad}(4)+1=4^2+2+1=19$ that also is square-free because is a prime number.
3) For $n=19$ also $19^{\operatorname{rad}(19)}+\operatorname{rad}(19)+1=19^{19}+19+1$ is a square-free integer since has prime factorization $3\cdot 139225573\cdot 4736724757839121$.
Computational evidence. Upto $N=50$ the only integers $m's$ such that $1\leq m\leq 50$ for which $m^{\operatorname{rad}(m)}+\operatorname{rad}(m)+1$ has some repeated prime factors are $m=13,20,22$ and $m=31$.
Too long for a comment : Checking the primes upto $10^6$, for the following $166$ n's , we do NOT get a squarefree number :