Rational solutions

Here's my solution: let $u = 2^x, v = 3^x$ so that the original equation becomes the quadratic:

$(2^x)^2 + 2^{x}3^{x} - (3^x)^2 = 0 \Rightarrow u^2 + uv - v^2 = 0$

which solving for $u$ becomes $u = \frac{-v \pm \sqrt{v^2 - 4(1)(-v^2)}}{2} = (\frac{-1 \pm \sqrt{5}}{2})v.$ Since $u,v > 0$ we will only admit the positive root, or:

$u = (\frac{-1 + \sqrt{5}}{2})v \Rightarrow (\frac{2}{3})^x = \frac{-1 + \sqrt{5}}{2} \Rightarrow x = \frac{ln(\sqrt{5}-1) - ln(2)}{ln(2) - ln(3)} = \boxed{-1 - \frac{ln(3) + ln(\sqrt{5}-1)}{ln(2) - ln(3)}}.$

Hence, there are no solutions for $x \in \mathbb{Q}.$

Relevant wiki: Fermat's Last Theorem

Because of Fermat's last theorem, x is less than 3. Inserting 1 or 2 makes the equation still false, meaning there isn’t any answer