Solve the system $3^x-2^{y^2}=77$, $3^{\frac{x}{3}}-2^{\frac{y^2}{2}}=7$ in $\mathbb{R}$ I had to solve the similar system $3^x-2^{y^2}=77,\; 3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7$ before, which can be solved like this:
$$3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7 \implies 2^{\frac{y^2}{2}}=3^{\frac{x}{2}}-7$$
$$2^{\frac{y^2}{2}}=3^{\frac{x}{2}}-7 \implies 2^{y^2}=(3^{\frac{x}{2}}-7)^2$$
$$2^{y^2}=(3^{\frac{x}{2}}-7)^2 \implies 2^{y^2}=3^{x}-14\cdot3^{\frac{x}{2}}+49$$
So putting that in the first equation:
$$3^x-2^{y^2}=77 \implies 3^x-3^{x}+14\cdot3^{\frac{x}{2}}-49=77$$
$$\mathbf{3^x-3^{x}+14\cdot3^{\frac{x}{2}}-49=77 \implies 14\cdot3^{\frac{x}{2}}=126}$$
$$14\cdot3^{\frac{x}{2}}=126 \implies 3^{\frac{x}{2}}=9$$
$$3^{\frac{x}{2}}=9 \implies x = 4$$
And putting the value of $x$ in the second equation we get $y$:
$$3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7 \implies 3^{2}-2^{\frac{y^2}{2}}=7$$
$$3^{2}-2^{\frac{y^2}{2}}=7 \implies 2^{\frac{y^2}{2}}=2$$
$$2^{\frac{y^2}{2}}=2 \implies y^2 = 2$$
$$y^2 = 2 \implies y = \sqrt{2}$$
So I tried solving the system in the question title with the same strategy until the step in bold, which gave me $3^x - 3^{\frac{2x}{3}}+14\cdot3^{\frac{x}{3}}=126$, and I have no idea how to progress further.

If $u=3^{\frac x 2}$ and $v=2^{\frac {y^2}2}$ then
$$\left \{
\begin{split}
u^3-v^2&=77\\
u-v&=7
\end{split}
\right.$$
So $v=u-7$ and thus $u^3-u^2+14u-126=0$.
Solving the latter gives you 3 roots (1 real and 2 complex), and you can find $u$ and $v$, and get back to $x$ and $y$.
