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
Hint If $P \neq 0$, then $P$ has the form $$P(x) = a_n x^n + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0 ,$$ where $a_n \neq 0$. Comparing the leading terms of $P(x^2)$ and $P(x)^2$ gives that $a_n = 1$. If $P(x) \neq x^n$, then there is some largest index $m < n$ such that $a_m \neq 0$, and so $P$ has the form $$P(x) = x^n + a_m x^m + O(x^{m - 1}) .$$ Now, substitute in both sides of the condition.
Incidentally, this argument works over any field of characteristic $\neq 2$ (in characteristic $2$, $P(x^2) = P(x)^2$ holds for all polynomials).