Repeated prime factors

Question

For $n\geqslant3$ integer, do all numbers in the form $$n^2-3n+1$$ have unique prime factors? In other words, if these numbers are written in prime factor form, are all indices of the primes in the form $1$?

I am not sure if this is true,. If this is actually true, is there a proof for it?

What I have done:

Let's first let $\ f(n) = n^2-3n+1$.
For example, $f(64)=3905$. Amazingly, $3905$ has primes $5$, $11$ and $71$, which corresponds to $f(4)$, $f(5)$ and $f(10)$. Since $f(n)$ lies between $(n-2)^2$ and $(n-1)^2$ for $n\geqslant4$, all $f(n)$ are not perfect squares. So, if it is possible to prove that all composite $f(n)$ have their prime factors from prime $f(n)$ (which is quite possible), this question is done.

Moreover can I generalise that: If any polynomial with integral coefficients $p(n)$ suffices that $p(n)$ is not the power of an integer for all integers $n$, then $p(n)$ in prime factor form cannot have any repeated prime factors?

Answer 1

I reckon that for $n=38$, $$n^2-3n+1=1331=11^3.$$