Can someone please check if I provided a correct and adequate proof to the following:
There are no positive integer solutions $(x, y, z)$ to the equation $$2x^6+3y^6=z^3 \tag{1}$$.
Here's how I approached the problem:
First suppose $(x, y, z)$ is a solution to $(1)$ and the $gcd(x, y, z) = d$, which means that $x=dx_1, y=dy_1, z=dz_1$, where $gcd(x_1, y_1, z_1) = 1$ and $d$ is an integer.
Substituting in and rearranging we get $$d^3(2x_1^6+3y_1^6)=z_1^3.$$
Upon dividing both sides by $d^3$ we obtain $$2x_1^6+3y_1^6=k^3,$$ where $k$ represents the quotient of $z_1$ and $d$.
We have obtained an equation very similar to $(1)$ with a "smaller" solution $(x_1, y_1, k)$, which yields a contradiction by Fermat's method of infinite descent.
Thus we have proven the statement.
Suggestions of all sorts are much appreciated.
Thanks in advance
As suggested in several question comments, e.g., this one, the main issue with your proof is that if $d = 1$, then your final equation is the same as the first. You haven't proven that $d \gt 1$. To show this, consider the equation modulo $7$. By Fermat's little theorem, if $7 \not\mid n$, then $n^6 \equiv 1 \pmod{7}$. This means $x^6$ and $y^6$ can only be congruent to $0$ or $1$ modulo $7$, so $2x^6 + 3y^6$ can only be congruent to $0$, $2$, $3$ or $5$ modulo $7$.
However, $z^3 \pmod{7}$ is equivalent to only one of $0$, $1$ or $6$. The only match with the LHS is when it's $0$, which means $x$, $y$ and $z$ must all be a multiple of $7$. Thus, you have $7 \mid d$, so $d \gt 1$ and then your argument of infinite descent will work. I'll leave it to you to update your proof accordingly.