The question is equivalent (by replacing $x$ with $\frac{1}{x}$) to asking about the irreducibility of $A(x) = x^{2k+1}-7x^{2k-1}+1 = x^{2k-1}(x^2-7)+1$ over $\mathbb{Q}$. An application of Rouché's theorem shows that $A(x)$ has exactly $2k-1$ zeroes (counted with multiplicity) of absolute value less than $1$. There remain two real zeroes $\beta_1 < -1$ and $\beta_2 > 1$. One can deduce from the equation they satisfy that $|\beta_i - (-1)^i\sqrt{7}| \leq \frac{1}{\sqrt{7}+1}$ ($i=1,2$).

Now suppose that $A$ factorizes as $A = BC$ with monic integer polynomials $B, C$. Wlog $B(\beta_1) = 0$. If $B(\beta_2) \neq 0$, then $$\beta_1^{2k-1}(\beta_1+\sqrt{7}) = \frac{1}{\sqrt{7}-\beta_1}$$ is an algebraic integer, all conjugates of which are less than $1$ in absolute value (the right hand side is less than $\frac{1}{2\sqrt{7}-\frac{1}{\sqrt{7}+1}}<1$ and all its conjugates are of absolute value at most $\frac{1}{\sqrt{7}-1}<1$).

So it would have to be $0$, which is absurd. So $B(\beta_2)=0$. But now all roots of $C$ are less than $1$ in absolute value, while its constant term is $1$. It follows that $C=1$ is constant and so $A$ is irreducible.

EDIT: I'm afraid that this answer is wrong and I hope you have some way of "unaccepting" it. The reason is that when taking conjugates of $\frac{1}{\sqrt{7}-\beta_1}$ we're allowed to freely change the sign of $\sqrt{7}$ (unless $\sqrt{7}$ is contained in the field generated by $\beta_1$, which seems rather unlikely). And $\frac{1}{-\sqrt{7}-\beta_1}$ is greater than $1$ in absolute value. I found this mistake only when I noticed that my "proof" gives that $x^4-7x^2+1=(x^2+3x+1)(x^2-3x+1)$ is irreducible. It should work with an integer square in place of $7$. Maybe someone can make it work also for $7$? I apologize for raising false hopes!

The question is equivalent (by replacing $x$ with $\frac{1}{x}$) to asking about the irreducibility of $A(x) = x^{2k+1}-7x^{2k-1}+1 = x^{2k-1}(x^2-7)+1$ over $\mathbb{Q}$. An application of Rouché's theorem shows that $A(x)$ has exactly $2k-1$ zeroes (counted with multiplicity) of absolute value less than $1$. There remain two real zeroes $\beta_1 < -1$ and $\beta_2 > 1$. One can deduce from the equation they satisfy that $|\beta_i - (-1)^i\sqrt{7}| \leq \frac{1}{\sqrt{7}+1}$ ($i=1,2$).

Now suppose that $A$ factorizes as $A = BC$ with monic integer polynomials $B, C$. Wlog $B(\beta_1) = 0$. If $B(\beta_2) \neq 0$, then $$\beta_1^{2k-1}(\beta_1+\sqrt{7}) = \frac{1}{\sqrt{7}-\beta_1}$$ is an algebraic integer, all conjugates of which are less than $1$ in absolute value (the right hand side is less than $\frac{1}{2\sqrt{7}-\frac{1}{\sqrt{7}+1}}<1$ and all its conjugates are of absolute value at most $\frac{1}{\sqrt{7}-1}<1$).

So it would have to be $0$, which is absurd. So $B(\beta_2)=0$. But now all roots of $C$ are less than $1$ in absolute value, while its constant term is $1$. It follows that $C=1$ is constant and so $A$ is irreducible.