Do bounded totally real integer polynomials with few level‑crossings force algebraic limits?

Motivation. For $m\ge1$, the equation $$\prod_{k=1}^n\Bigl(x+\cos^m\frac{k\pi}{n}\Bigr)=1$$ has a unique solution $x_n>0$, and $x_n\to\alpha$ with $\alpha$ algebraic: since $\frac1n\log\prod_k(\alpha+\cos^m\tfrac{k\pi}n)\to\frac1\pi\int_0^\pi\log|\alpha+\cos^m\theta|\,d\theta$, the limit is a zero of a logarithmic Mahler measure, hence algebraic by Jensen's formula (e.g. $\alpha=5/4$ for $m=1$, $9/16$ for $m=2$). The proof uses everything about the roots: they are $-\cos^m(k\pi/n)$, so $\frac1n\log|P_n(x)|$ converges to an explicit integral over $[0,\pi]$, which is a logarithmic Mahler measure, and Jensen's formula gives algebraicity. Nothing in this argument works if the roots are merely some bounded real algebraic integers with no formula. The question is what happens when the explicit structure of the roots is removed. The cosine family has the coarse features — monic in $\mathbb Z[x]$, all roots real in a fixed interval, no high multiplicity, the level set crossed only boundedly often — but at the exponential level $Q(n)=2^{nm}$ those features are not what makes the limit algebraic: the Chebyshev product below has the same coarse features with $Q(n)=3^n$ and a transcendental limit. So at exponential level the answer depends on the fine structure of the roots, and Jensen's formula is what settles the cosine case. Conjecture. Let $Q:\mathbb N\to\mathbb N$ satisfy $\log Q(n)=o(n)$. Suppose $P_n\in\mathbb Z[x]$ is monic of degree $n$, every root of $P_n$ is real and lies in a fixed bounded interval, and both the multiplicities of the roots of $P_n$ and the number of real solutions of $P_n(x)=Q(n)$ are bounded independently of $n$. Then every convergent sequence $x_n$ with $P_n(x_n)=Q(n)$ has algebraic limit. Since $P_n(x)-Q(n) \in \mathbb{Z}[x]$ this means each $x_n$ is an algebraic integer with degree at most $n$ but the limit of a convergent sequence of algebraic integers need not be algebraic. To show that $\log Q(n)=o(n)$ is necessary, I will produce a counterexample for $Q(n) = 3^n$. Let$$P_n(x) = U_{\lfloor n/\sqrt{2} \rfloor}(x/2)U_{n-\lfloor n/\sqrt{2} \rfloor}((x-1)/2)$$where $U_i(x)$ is the Chebyshev polynomial of the second kind. This is a monic polynomial in $\mathbb{Z}[x]$ where each root has multiplicity at most $2$ and all the roots of $P_n(x)$ lie in $(-2,3)$. It can be shown that $P_n(x) = 3^n$ has at most two solutions. Now$$P_n(x)^{1/n} \to \left(\dfrac{x+\sqrt{x^2-4}}{2}\right)^{1/\sqrt{2}} \left(\dfrac{x-1+\sqrt{(x-1)^2-4}}{2}\right)^{1-1/\sqrt{2}}$$and if $x_n \to \alpha$ then we have$$\left(\dfrac{\alpha+\sqrt{\alpha^2-4}}{2}\right)^{1/\sqrt{2}} \left(\dfrac{\alpha-1+\sqrt{(\alpha-1)^2-4}}{2}\right)^{1-1/\sqrt{2}} = 3.$$This means that$$\left(\dfrac{\alpha+\sqrt{\alpha^2-4}}{\alpha-1+\sqrt{(\alpha-1)^2-4}}\right)^{1/\sqrt{2}} = \dfrac{6}{\alpha-1+\sqrt{(\alpha-1)^2-4}}.$$If $\alpha$ was algebraic then so would $\frac{\alpha+\sqrt{\alpha^2-4}}{\alpha-1+\sqrt{(\alpha-1)^2-4}}$ and since the ratio is more than $1$ and $1/\sqrt{2}$ is algebraic and irrational it means the left hand side is transcendental by Gelfond-Schneider while the right hand side is algebraic, which is a contradiction. Therefore $\alpha$ is transcendental. Is this conjecture known, or a consequence of known results (Fekete–Szegő, Bilu, Schur–Siegel–Smyth and Smith's work on limit measures of totally real algebraic integers)?