Find all $x > 0$ s/t $\sqrt{x} + 1/\sqrt{x}$ and $x^{1/3} + 1/x^{1/3}$ are integers. Why doesn't this get all solutions? This is a math puzzle I've been working on, and I'm not sure why my approach isn't yielding all the solutions for $x$. As in the title, the question is to find all positive reals $x$ where $\sqrt{x} + 1/\sqrt{x}$ and $x^{1/3} + 1/x^{1/3}$ are both integers. Here's my approach--I'm not sure where the error is. Where does the following approach go wrong?
Suppose that $\sqrt{x} + 1/\sqrt{x} = m$ and $x^{1/3} + 1/x^{1/3} = n$, where $m,n \in \mathbb{Z}$. We let $m, n > 0$ since $\sqrt{x} + 1/\sqrt{x} > 0$ and $x^{1/3} + 1/x^{1/3} > 0$ for $x > 0$.
We equivalently have $x - m\sqrt{x} + 1 = 0$ and $x^{2/3} -nx^{1/3} + 1 = 0$. Let $u = \sqrt{x}$, so that we have $u^2 - mu + 1 = 0$ and $u^{4/3} - nu^{2/3} + 1 = 0$. It suffices to find all $u$ satisfying both of these equations.
Applying the quadratic formula, $u = \frac{m \pm \sqrt{m^2 - 4}}{2}$ and $u^{2/3} = \frac{n \pm \sqrt{n^2-4}}{2}$. Squaring the first equation gets $u^2 = (\frac{m \pm \sqrt{m^2 - 4}}{2})^2$.Cubing the second equation gets $u^2 = (\frac{n \pm \sqrt{n^2-4}}{2})^3$.
So there are 4 sign combinations setting these two expressions for $u^2$ equal. Since $m,n$ are integers, I look for integer solutions only, finding only $m = 2, n = 2$ as the viable solution in all 4 cases. I checked this part with WolframAlpha--the general idea was to isolate radicals to one side. If we have a nonzero sum of surds, it ends up being irrational while the other side is rational. That leaves integer solutions to the radicals $\sqrt{m^2-4},\sqrt{n^2-4}$, the only positive solutions to which are $m = 2$ and $n = 2$.
This yields $u = 1$ as the only viable solution. However, the solution states that there are infinitely many such $x$.

A good way to simplify algebra is to start by getting rid of the radicals.  Let $y=\sqrt[6]x$.  Then $\sqrt{x}=y^3$ and $\sqrt[3]x=y^2$ and we have to find such $y$ where
$$
y^3+1/y^3 = m
$$
$$
y^2+1/y^2 = n
$$
Further, let $a=y+1/y$.  Then $$n = a^2 -2$$ and $$m = a^3-3a = a(a^2-3) = a(n-1)$$
Therefore, $a$ must also be an integer ($a=\frac{m}{n-1}$ is rational and its square is an integer $a^2=n+2$)
Solving $a=y+1/y$ for $y$, we get
$$y=\frac{a\pm\sqrt{a^2-4}}{2}$$ and
$$x=y^6 = \left({\frac{a\pm\sqrt{a^2-4}}{2}}\right)^6$$ for any integer $a\ge2$
