Solve for $x$ if $4^{\frac{x}{y} + \frac{y}{x}}$ $=$ $32$ and $\log_3(x+y)+\log_3(x-y)=1$ 

Question:
Solve for $x$ if $4^{\frac{x}{y} + \frac{y}{x}}$ $= 32$ and $\log_3(x+y)+\log_3(x-y)=1$



My attempt: With the first equation
$$4^{\frac{x}{y} + \frac{y}{x}} = 32$$
$$2^{2(\frac{x}{y} + \frac{y}{x})} = 2^5$$
$$ 2(\frac{x}{y} + \frac{y}{x}) = 5 $$
$$ \frac{x}{y} + \frac{y}{x} = \frac{5}{2} $$
$$ \frac{x^2 + y^2}{xy} = \frac{5}{2} $$
Now with the second equation
$$\log_3(x+y)+\log_3(x-y)=1$$
$$\log_3((x+y)(x-y)) = 1 $$
$$ \log_3(x^2-y^2) = 1 $$
$$ x^2-y^2 = 3$$
Now I have 2 equations:
$$ \frac{x^2 + y^2}{xy} = \frac{5}{2} $$
$$ x^2-y^2 = 3$$
Now I am stuck..

You have the two equations $$\frac{x^2 + y^2}{xy} = \frac{5}{2}\tag 1$$ $$x^2-y^2 = 3\tag 2$$ From $(2)$, $y=\pm \sqrt{x^2-3}$. So, plugging in $(1)$ $$\frac{x^2 + y^2}{xy} = \frac{5}{2}\implies 2(x^2+y^2)=5 x y \implies 2(2x^2-3)=\pm 5x\sqrt{x^2-3}\tag 3$$ Square both sides $$4(2x^2-3)^2=25x^2(x^2-3)$$ Expand and group terms to get $$-9 x^4+27 x^2+36=0$$ which is a quadratic in $t$ if $t=x^2$; its roots are $-1$ and $4$.
I am sure that you can take it from here.
