I think I got the proof that no such real polynomial with degree $ \geq 6$ exists.
Let $n \geq 6$
Suppose for contradiction that $z_1,\ldots,z_n \in \mathbb R-\{0\}^n$ are such that $(X-z_1)...(X-z_n)=X^n+\sum_{k=1}^{n-1}z_iX^{n-i}$
Then three useful identities appear $$\sum_{k=1}^{n}z_k=-z_1 \; \; \; \;(1)$$
$$\sum_{\large1\leq i<j \leq n}z_iz_j=z_2 \; \; \; \;(2)$$
$$\prod_{k=1}^n z_k=(-1)^n z_n \; \; \; \;(3)$$
Since $$(\sum_{k=1}^{n}z_k)^2=\sum_{k=1}^{n}z_k^2+2\sum_{\large1\leq i<j \leq n}z_iz_j$$it follows that$$z_1^2=2z_2+\sum_{k=1}^{n}z_k^2$$
Hence $$0< \sum_{k=2}^{n}z_k^2=-2z_2 \; \; \; \;(4)$$ and $$0<\sum_{k=3}^{n}z_k^2=1-(z_2+1)^2 \; \; \; \;(5) $$
$(4)$ and $(5)$ imply $$\; \; \; \;-2<z_2<0 \; \; \; \;(6)$$
thus $(6)$ and $(4)$ imply $$0<\sum_{k=2}^{n}z_k^2 < 4 \; \; \; \; (7)$$
Also $(6)$ and $(5)$ imply $$0<\sum_{k=3}^{n}z_k^2 \leq 1 \; \; \; \; (8)$$
By AM-GM, $$\left(|z_3|^2\ldots|z_{n-1}|^2 \right)^{1/(n-3)} \leq \frac{1}{n-3}\sum_{k=3}^{n-1}z_k^2 \leq \frac{1}{n-3}\sum_{k=3}^{n}z_k^2$$
Hence
$$|z_3|^2\ldots|z_{n-1}|^2 \leq \left(\frac{1}{n-3}\sum_{k=3}^{n}z_k^2\right)^{n-3} $$
Squaring, $$|z_3|\ldots|z_{n-1}| \leq \left(\frac{1}{n-3}\sum_{k=3}^{n}z_k^2\right)^{\large \frac{n-3}{2}} \leq_{ \large (8)} \dfrac{1}{{(n-3)}^{(n-3)/2}} \; \; \; \; (9)$$
By triangle inequality $(1)$, and Cauchy-Schwarz
$$2|z_1| \leq \sum_{k=2}^{n}|z_k| \leq \sqrt{n-1} \sqrt{\sum_{k=2}^{n}z_k^2} $$
Hence by $(7)$,
$$|z_1| \leq \sqrt{n-1} \; \; \; \; (10)$$
Rewriting $(6)$ as $$|z_2|\lt2 \; \; \; \; (11) $$
Recalling $(3)$ (with $z_n$ cancelled from both sides) and putting together $(9)$, $(10)$ and $(11)$, we have
$$1=|z_1||z_2||z_3|\cdots|z_{n-1}| < \dfrac{ 2\sqrt{n-1}}{{(n-3)}^{(n-3)/2}}$$
This inequality fails for $n\geq 6$.
Contradiction.
I can't prove anything for $n=5$ so maybe the conjecture doesn't hold.
Part 1 is still kind of nice; depending on what you need this for this may or may not be useful:
For a real polynomial, its roots are either real or form conjugate pairs. If you have two conjugate roots of unity
$$e^{\frac{2i\pi k}{n}},\ e^{-\frac{2i\pi k}{n}},$$
then the quadratic polynomial with them as roots is
$$P(x)=x^2-2\cos\left(\frac{2\pi k}{n}\right)x+1.$$
So, any real-coefficiented polynomial (whose roots are all roots of unity) has the form
$$(x+1)^m(x-1)^n\prod_{k=1}^N \left(x^2-2\cos\left(\frac{2\pi a_k}{b_k}\right)x+1\right),$$
where $m,n$ are nonnegative integers and all $a_k,b_k$ are positive integers (with $a_k<b_k$, if you wish).
For part 2, I mean, you can just pick the roots to be
$$x_k=e^{\frac{2i\pi a_k}{b_k}},$$
and your polynomial will be
$$\prod_{k=1}^N \left(x-e^{\frac{2i\pi a_k}{b_k}}\right),$$
but this probably isn't that useful. Another thing that may or may not be useful for either of these is that all polynomials with only roots of unity as roots have roots that all satisfy $x^M=1$ for some (possibly large with respect to the degree of the polynomial) integer $M$. In the case above this $M$ can be taken to be $b_1b_2\cdots b_k$. So, all polynomials with only roots of unity as roots are factors of $x^M-1$ for some $M$.
Best Answer
I'll assume you want the divisibility to be in $\mathbb{Q}[x]$. Write $\displaystyle f(x) = \prod_{i=1}^{k} (x - a_i)^{m_i}$ where the $a_i$ are distinct (in $\mathbb{C}$).
Lemma: $\displaystyle \frac{f'(x)}{f(x)} = \sum_{i=1}^{k} \frac{m_i}{x - a_i}$.
This is a direct consequence of the product rule and you do not need to use any calculus to prove it. (That is, you don't need to take logarithms. The derivative can be defined as a totally formal operation on polynomials and this identity holds in $\mathbb{Q}(x)$.)
If $f'(x)$ divides $f(x)$, then $\frac{f'(x)}{f(x)} = \frac{n}{x - a}$ for some $a$. Now, the functions $\frac{1}{x - a}$ are linearly independent (you also do not need to use any calculus to prove this) in $\mathbb{C}(x)$. It follows that this is possible if and only if $f$ has $a$ as a repeated root of multiplicity $n$, so $a$ is the only root.
It follows that $f$ has the form $c(px - q)^n$ for some integers $c, p, q$ with $a, p \neq 0$. (I'm ignoring the degenerate cases.)