<s>[INST] On the conjecture that if $\,\forall\{x, y, z\}\subset \Bbb{Z^+}$ squarefree, and $\ x^3 + y^3 = z^2$, then $\gcd(3, x, y) = 1$. I made a conjecture that if $\,\forall\{x, y, z\}\subset\mathbb{Z}^+$ with $x, y, z$ squarefree (not raised to a power greater than $1$; not a perfect power), and $$x^3 + y^3 = z^2,$$ then $\gcd(3, x, y) = 1$. I was able to prove my conjecture for all $z$ prime.

Proof: Let $z$ be a prime number.
Since $(x+y)^3 = x^3 + y^3 + 3xy(x+y),$ then $$\begin{align} x^3 + y^3 &= (x+y)^3 - 3xy(x+y) \\ &= (x+y)\big((x+y)^2-3xy\big) \\ &=(x+y)(x^2 + y^2 - xy).\tag1\end{align}$$ Therefore, we need integer solutions such that $(*)\,\, x + y = x^2 + y^2 - xy$ because $(1) = z^2$, follows that $$\begin{align} 0 &= x^2 + y^2 - xy - x - y \\ &= x(x - 1) + y(y - 1) - xy \\ \Leftrightarrow 3xy &= x(x-1) + y(y-1) + 2xy.\end{align}$$ Since $3xy$ is not a square number, this means that $x \neq 3y$, $y\neq 3x$ and $3\neq xy$.
Therefore $\gcd(3, x, y) = 1$ for this case, but there is also another case where $(**)\,\, x + y = z$ and $x^2 + y^2 - xy = 1$. Of course $x + y\neq 1$ because $x, y\in\mathbb{Z}^+$, thus we need to consider statement $(**)$. $$\begin{align} x^3 + y^3 &= (x+y)^2 \\ \Leftrightarrow (x+y)(x^2 + y^2 - xy) &= (x+y)^2 \\ \Leftrightarrow x^2 + y^2 - xy &= x + y\tag*{$\unicode{x21af}$ (Contradiction)}\end{align}$$ Therefore, statement $(*)$ must always be true, i.e. $\gcd(3, x, y) = 1$.$\quad\quad\quad\quad\qquad\qquad\qquad\quad\Box$

Of course if $z$ is not prime, then $x+y$ does not necessarily have to equal $x^2 + y^2 - xy$. Take $z = 6$ for example, then $z^2 = 6^2 =6\times 6$ but it also equals $9\times 4$, etc.
Could somebody prove/disprove my conjecture for all $z\in\mathbb{Z^+}$? Along with that, is my proof valid?
Thank you in advance.

Edit:
The symbol $\unicode{x21af}$ (Harry Potter's Scar) is used to symbolise a contradiction in this case. Most people prefer $\Rightarrow\Leftarrow$, so if you prefer that instead, treat the Scar as if it were the arrows.
 [/INST] As stated your conjecture is false. 
I supposed that $x,y$ are squarefree integers.
The following paramertization holds,
$$ x = s^{4} + 6 \, s^{2} t^{2} - 3 \, t^{4}, \ 
y = -s^{4} + 6 \, s^{2} t^{2} + 3 \, t^{4},\  
z =  6 \, {\left(s^{4} + 3 \, t^{4}\right)} s t$$
(See for instance lemma 3.2.6 of Zagier http://www.cecm.sfu.ca/~nbruin/thesis.pdf).
Then we get many solutions with $\gcd(x,y,3)>1$ and $x,y,z>0.$
For instance $s=-3,t=2$ provides $(x,y,z)=(249, 183, 4644)$
or $s=-3,t=4$ gives $(x,y,z)=(177, 1551, 61128).$
In your proof, there are more conditions except $x+y=x^2+y^2-xy$ that implies $(1)=z^2$, e.g. $x+y = ab, x^2+y^2-xy=abc^2.$
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