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..

$$\frac{x^2+y^2}{xy}=\frac 52$$
$$y=xt$$
$$\frac {x^2+x^2t^2}{x^2t}=\frac 52$$
$$\frac{1+t^2}{t}=\frac 52$$
$$t=2; t=\frac 12$$
$y=2x \Rightarrow x^2-4x^2=3 x^2-4x^2=3 \Rightarrow x^2=-1$
$y=\frac 12x \Rightarrow x=2y \Rightarrow 4y^2-y^2=3 \Rightarrow y=\pm 1; x =\pm 2$
