Carmichael number square free

We start as in your post, letting $n=p^tm$ where $t\ge 2$ and $m$ is not divisible by $p$.

By the Chinese Remainder Theorem, the system of congruences $x\equiv 1+p\pmod{p^t}$, $x\equiv 1\pmod{m}$ has a solution $a$. Note that $\gcd(a,n)=1$.

Since $n$ is Carmichael, we have $a^{n-1}\equiv 1\pmod{n}$. In particular, $a^{n-1}\equiv 1\pmod{p^t}$, and therefore $a^{n}\equiv a\pmod{p^2}$.

So $(1+p)^{n}\equiv 1+p\pmod{p^2}$. Expand $(1+p)^{n}$ modulo $p^2$ using the binomial theorem. We get that $(1+p)^{n}\equiv 1\pmod{p^2}$, since the first two terms of the expansion are $1$ and $np$, and the rest of the terms are divisible by $p^2$.

Thus $1\equiv 1+p\pmod{p^2}$. This is impossible.

You ask about a part of the theorem of Korselt of which I adjoint a copy. Unfortunately I don't remember the source of the publication however I give you a curious detail: Korselt has made his discovery before these numbers were called Carmichael and could not find any. It was Carmichael who discovered the first, 561,several years after this theorem.