Short Answer: the polynomials
$$P_{2n+1}(x) = x^{2n+1}(x^8 - x^7 - x^6 + x^4 - x^3 + x + 1) - (x^8 + x^7 - x^5 + x^4 - x^2 - x + 1)$$
for $n \ge 7$ should have an irreducible factor with exactly two roots $\alpha_n$ and $\beta_n$ of modulus greater than $1$, and $\alpha_n - \beta_n$ is exponentially converging to zero. The irreducibility of the non-cyclotomic factor is a consequence of Lehmer's conjecture, but can probably be established by direct elementary means if one wished to do so.

This construction is a little elaborate, so I give details. Let

$$f(x) = x^8 - x^7 - x^6 + x^4 - x^3 + x + 1$$

This polynomial is carefully chosen so that it has the following properties. First, it is non-reciprocal, that is, it is distinct from

$$f^*(x):= x^8 f(1/x) = x^8 + x^7 - x^5 + x^4 - x^2 - x + 1.$$

Second, it factors into a product of cyclotomic polynomials times the *square* of an irreducible polynomial whose root is a Pisot number,
that is,

$$f(x) = (x^2 - x + 1)(x^3 - x - 1)^2.$$

The first polynomial is cyclotomic, the second has a unique root

$$\alpha \sim 1.32472\ldots$$

of absolute value greater than one and two complex conjugate roots inside the unit circle. Now let

$$P_n(x) = f(x) x^n - f^*(x) = f^*(x) \left( \frac{f(x)}{f^*(x)} x^n - 1 \right),$$

For $n > 8$, this is a polynomial with coefficients either $-1$, $0$, or $1$. An argument (due to David Boyd in the 70s, see his Duke paper) shows that this construction gives a polynomial with precisely $2$
roots outside the unit circle for $n$ large enough ($n \ge 15$ will suffice in this case). Moreover, the two roots $\alpha_n$ and $\beta_n$
will have the property that

$$\lim_{n \rightarrow \infty} \alpha_n = \lim_{n \rightarrow \infty} \beta_n = \alpha,$$

so in particularly $|\alpha_n - \beta_n|$ is not bounded below, answering your question in the negative. In fact, the difference converges to zero very fast, the difference is of the order $\alpha^{-n/2}$ up to some constant.

It remains to consider irreducibility up to cyclotomic factors.
Actually, when $n$ is even, the polynomial is reducible, because --- up to a factor of $(x^2-x + 1)$, it is a difference of two squares, since

$$P_n(x) = (x^2 - x + 1)((x^3 - x - 1)^2 x^n - (x^3 + x^2 - 1)^2).$$

So we want to concentrate on $P_{2n+1}(x)$.
For convenience, write $P_n(x) = (x^2 - x + 1) Q_n(x)$.
I suspect that $Q_n(x)$ (and so $P_n(x)$) is irreducible for all odd $n$ up to cyclotomic factors (which will only depend on $n \bmod 30$ by a non-trivial but somewhat standard computation related to vanishing sums of roots of unity), but this may well be tedious to prove. It is not immediately apparent how to do this, but I haven't spent too long trying to do so. Hopefully you will be content with a proof of irreducibility **assuming** Lehmer's conjecture that any polynomial has Mahler measure at least

$$\eta = 1.17628\ldots $$

where $\eta$ is a root of Lehmer's degree $10$ polynomial

$$x^{10} + x^{9} - x^{7} - x^{6} - x^{5} - x^{4} - x^{3} + x + 1 = 0.$$

Any non-cyclotomic monic polynomial has at least one root of absolute value greater than one by Kronecker. For any $n$, there are at most two non-cyclotomic factors of $Q_n(x)$, since there are only two roots of absolute value more than $1$. Write $Q_{2n+1}(x) = A(x) B(x) \Phi(x)$ where $\Phi(x)$ is the cyclotomic factor and $A(x)$ and $B(x)$ are irreducible. Then

$$ \Phi(x^2) A(x^2) B(x^2) =((x^6 - x^2 - 1)^2 x^{4n+2} - (x^6 + x^4 - 1)^2)$$
$$ = ((x^6 - x^2 - 1) x^{2n+1} - (x^6 + x^4 - 1))((x^6 - x^2 - 1) x^{2n+1} + (x^6 + x^4 - 1)).$$
$$ = - R(x) R(-x).$$

Any common factor of $R(x)$ and $R(-x)$ must divide their sum and their difference, but

$$((x^6 - x^2 - 1) x^{2n+1}, x^6 + x^4 - 1) = 1,$$

so they have no common factor. Note that $A(x^2)$ and $B(x^2)$ cannot have cyclotomic factors (although they may be reducible). Suppose that $A(x^2)$ was irreducible. Then if $A(x^2) | R(x)$, then $A(x^2) = A((-x)^2)$ would divide $R(-x)$, which is a contradiction. Hence both $A(x^2)$ and $B(x^2)$ must be reducible. Thus $P_{2n+1}(x^2)$ has four irreducible factors which are not cyclotomic. But the Mahler measure of $P_{2n+1}$ is very close to $\alpha^2$, so the Mahler measure of at least one of the four factors is bounded above by a factor which is very close to

$$\alpha^{1/2} \sim 1.150963\ldots < 1.17628\ldots = \eta,$$

where $\eta$ is Lehmer's number. But this would contradict Lehmer's conjecture. Hence Lehmer's conjecture implies that the non-cyclotomic part of $P_{2n+1}(x)$ is irreducible.