There exist irreducible monic polynomials such that all their roots apart from two lie on the unique circle (and are not roots of unity). Such polynomials can be chosen among Salem polynomials and they exit in arbitrary high degree. By definition a Salem polynomial $S(x)\in \mathbb Z[x]$ is a monic irreducible reciprocal polynomial with exactly two roots off the unit circle, both real and positive. Of course non of the roots of these polynomials are roots of unity, since these polynomials are irreducible.
See for example theorem 1.6 in the article
Automorphisms of even unimodular lattices and unramified Salem numbers of Gross and Mcmullen:
http://www.math.harvard.edu/~ctm/home/text/papers/unim/unim.pdf
Theorem. For any odd integer $n\ge 3$ there exist infinitely many unramified Salem polynomials of degree $2n$.
There is the following result of Davenport, Lewis, and Schinzel [DLS64, Cor to Thm 2]:
Theorem. Let $p \in \mathbf Z[x]$. Then the following are equivalent:
- $p$ is a sum of two squares in $\mathbf Z[x]$;
- $p(n)$ is a sum of two squares in $\mathbf Z$ for all $n \in \mathbf Z$;
- Every arithmetic progression contains an $n$ such that $p(n)$ is a sum of two squares in $\mathbf Z$.
Criterion 3 is really weak! For example, it shows that in 2, we may replace $\mathbf Z$ by $\mathbf N$. Because it's short but takes some time to extract from [DLS64], here is their proof, simplified to this special case.
Proof. Implications 1 $\Rightarrow$ 2 $\Rightarrow$ 3 are obvious. For 3 $\Rightarrow$ 1, factor $p$ as
$$p = c \cdot p_1^{e_1} \cdots p_r^{e_r}$$
with $p_j \in \mathbf Z[x]$ pairwise coprime primitive irreducible and $c \in \mathbf Q$. We only need to treat the odd $e_j$ (and the constant $c$). Let $P = p_1 \cdots p_r$ be the radical of $p/c$, and choose $d \in \mathbf N$ such that $P$ is separable modulo every prime $q \not\mid d$. Suppose $P$ has a root modulo $q > 2d\operatorname{height}(c)$; say
$$P(n) \equiv 0 \pmod q$$
for some $n$. Then $P'(n) \not\equiv 0 \pmod q$, hence $P(n+q) \not\equiv P(n) \pmod{q^2}$. Replacing $n$ by $n+q$ if necessary, we see that $v_q(P(n)) = 1$; i.e. there is a $j$ such that
$$v_q\big(p_i(n)\big) = \begin{cases}1, & i = j, \\ 0, & i \neq j.\end{cases}.$$
If $e_j$ is odd, then so is $v_q(p(n))$, which equals $v_q(p(n'))$ for all $n' \equiv n \pmod{q^2}$. By assumption 3 we can choose $n' \equiv n \pmod{q^2}$ such that $p(n')$ is a sum of squares, so we conclude that $q \equiv 1 \pmod 4$. If $L = \mathbf Q[x]/(p_j)$, then we conclude that all primes $q > 2d\operatorname{height}(c)$ that have a factor $\mathfrak q \subseteq \mathcal O_L$ with $e(\mathfrak q) = f(\mathfrak q) = 1$ (i.e. $p_j$ has a root modulo $q$) are $1$ mod $4$. By Bauer's theorem (see e.g. [Neu99, Prop. VII.13.9]), this forces $\mathbf Q(i) \subseteq L$.
Thus we can write $i = f(\theta_j)$ for some $f \in \mathbf Q[x]$, where $\theta_j$ is a root of $p_j$. Then $p_j$ divides
$$N_{\mathbf Q(i)[x]/\mathbf Q[x]}\big(f(x)-i\big) = \big(f(x)-i\big)\big(f(x)+i\big),$$
since $p_j$ is irreducible and $\theta_j$ is a zero of both. Since $f(x)-i$ and $f(x)+i$ are coprime and $p_j$ is irreducible, there is a factor $g \in \mathbf Q(i)[x]$ of $f(x)+i$ such that
$$p_j = u \cdot N_{\mathbf Q(i)[x]/\mathbf Q[x]}(g) = u \cdot g \cdot \bar g$$
for some $u \in \mathbf Q[x]^\times = \mathbf Q^\times$. Applying this to all $p_j$ for which $e_j$ is odd, we get
$$p = a \cdot N_{\mathbf Q(i)[x]/\mathbf Q[x]}(h)$$
for some $h \in \mathbf Q(i)[x]$ and some $a \in \mathbf Q^\times$. By assumption 3, this forces $a$ to be a norm as well, so we may assume $a = 1$. Write $h = \alpha H$ for $\alpha \in \mathbf Q(i)$ and $H \in \mathbf Z[i][x]$ primitive. Then
$$p(x) = |\alpha|^2 H \bar H,$$
so Gauss's lemma gives $|\alpha|^2 \in \mathbf Z$. Since $|\alpha|^2$ is a sum of rational squares, it is a sum of integer squares; say $|\alpha|^2 = |\beta|^2$ for somce $\beta \in \mathbf Z[i]$. Finally, setting
$$F + iG = \beta H,$$
we get $p = F^2 + G^2$ with $F, G \in \mathbf Z[x]$. $\square$
Footnote: I am certainly surprised by this, given that the version for four squares is clearly false. Indeed, the condition just reads $p(n) \geq 0$ for all $n \in \mathbf Z$. But the OP's example cannot be written as any finite sum of squares in $\mathbf Z[x]$, because exactly one of the terms can have positive degree. (However, it might be different in $\mathbf Q[x]$.)
References.
[DLS64] H. Davenport, D. J. Lewis, and A. Schinzel, Polynomials of certain special types. Acta Arith. 9 (1964). ZBL0126.27801.
[Neu99] J. Neukirch, Algebraic number theory. Grundlehren der Mathematischen Wissenschaften 322 (1999). ZBL0956.11021.
Best Answer
Yes. For nonnegative (as opposed to strictly positive) coefficients, a more general result is available (in several variables). See the article by me on powers of polynomials in Symbolic Dynamics and Its Applications (Editors, Adler and Walters, AMS 1992 (unfortunately, I can't find a link to it) [I have three articles there]. In several variables, the formulation is, if $P$ is a polynomial such that $P(1,1,\dots,1)> 0$, and there exists $m$ such that $P^m$ has only nonnegative coefficients, then for all sufficiently large $M$, $P^M$ has only nonnegative coefficients.
In one variable, the strict positivity condition you require can also be dealt with from this, but is a nonstarter in more than two variables. Note that if the second largest or the second smallest degree monomial has zero coefficient, then that persists in all powers, so no power of such poly in one variable can be strictly positive in your sense. However, the answer to your question is yes, because the gaps will eventually fill in ... (this requires a little, but not much, work).
Edit: Here is the reference (found, finally, in Zentralblatt): Handelman, David, Polynomials with a positive power. (Zbl 0794.26013) Symbolic dynamics and its applications, Proc. AMS Conf. in honor of R. L. Adler, New Haven/CT (USA) 1991, Contemp. Math. 135, 229-230 (1992). [For the entire collection see Zbl 0755.00019.]