This statement from your lecture notes
If a number $n>1$ is not prime, then it has a prime factor.
is true - but it's not a very good way to say what it's trying to say.
Here's an expanded version.
Every integer $n > 1$ has a prime factor. If $n$ happens to be prime then that prime factor is $n$ itself. If $n$ is not prime then it has a prime factor less than itself.
For the last part of your question: every prime is an integer, a natural number and a positive integer since every positive integer is also a natural number and an integer. It's probably best to use the most restrictive description - a prime is a positive integer - in fact, an integer greater than 1.
As pointed out in the comments:
If the question is asking for exactly $n$ factors, then the claim is false for every $n$, for the following simple reason:
If $p \neq 2,$ then $p \equiv 1 \pmod{2},$ thus $p-1$ and $p+1$ are both divisible by $2$, thus can not have different prime factorizations.
However, the statement is true for at least $n$ different prime factors:
Let $p_1,...,p_n, q_1,...,q_n, r_1,...,r_n$ be different odd prime numbers.
Let $P = p_1p_2 \cdots p_n, Q =q_1q_2\cdots q_n$ and $ R= r_1r_2\cdots r_n.$
(1) By the Chinese Remainder Theorem,
\begin{align*}
x-1 &\equiv 0 &\pmod{P} \\
x+1 &\equiv 0 &\pmod{Q}\\
x+2 &\equiv 0 &\pmod{R} \\
\end{align*}
has a solution.
(2) Let $x\in \mathbb{N}$ be a solution of the congruence equations. Note that $x_k = x+ k \cdot PQR $ is also a solution for all $k\in \mathbb{N}$.
(3) Now $(x,PQR) =1$, so by Dirichlet's theorem of primes in arithemetic progression you get the result.
Best Answer
Yes and yes. In fact, given any finite set of primes $p_i \equiv 1 \pmod 4$, there exists an integer $n$ such that $n^2 + 1$ is divisible by all of the $p_i$. Proof: each of the congruences $n_i^2 \equiv -1 \pmod{p_i}$ individually has two solutions modulo $p_i$; choose one of the two for each $i$, and use the Chinese Remainder Theorem to find an $n$ congruent to each $n_i$ modulo $p_i$.