Prove that $y^3=x^2+2^{16}3^9$ has no integer solutions.
Let $k=2^83^4,$ then $y^3=x^2+3k^2=(x+\sqrt{-3}k)(x-\sqrt{-3}k),$ but that's going to be complex.
I wonder is there a better way to solve this problem?
Edit: Sorry, this statement is NOT true, since $1728^3=62208^2+2^{16}3^9.$ I saw this problem from a website, but that OP gave me a false statement..
Let us first observe that, for $p=2$ or $3$, if $p\mid x$, then $p\mid y$, so $p^2\mid x$, hence $p^2\mid y,$ and $p^3\mid x.$
Therefore there are six cases, respectively given by the equations $y^3=x^2+k,$ with $k=2^{k_1}3^{k_2},$ where $k_1\in\{16,10,4\}, k_2\in\{9,3\}.$ In each of these, $x$ is relatively prime to both $2$ and $3$.
Now let $k'_1=k_1/2, k'_2=(k_2-1)/2$ and write $y^3=(x-k'\sqrt{-3})(x+ik'\sqrt{-3}),$ where $k'=2^{k'_1}3^{k'_2}.$
Now, by means of Minkowski bound, or of any other standard calculations, we know that the class-number of $\mathbb Q(\sqrt{-3})$ is $1$. Thus, by our assumptions, we know that there are $a,b,c,d\in\mathbb Z$ such that, setting $\alpha=\frac{1+\sqrt{-3}}{2}, \beta=a+b\alpha, \gamma=c+d\alpha,$ we have that $y=\beta\gamma, x-k'\sqrt{-3}=\epsilon\beta^3, x+k'\sqrt{-3}=\epsilon'\gamma^3,$ where $\epsilon,\epsilon'$ are units.
Now the unit group of $\mathbb Q(\sqrt{-3})$ is a cyclic group of order $6$ generated by $\alpha.$ So all units are powers of $\alpha$; say $\epsilon=\alpha^m$.
Expanding out the equations, we find that $-k'=\pm\begin{cases}3ab(a+b)/2\\(a^3-b^3+3a^2b)/2\\(a^3-b^3-3ab^2)/2\end{cases}.$
Hence our problem now is to examine closely these $18$ cases.
Indeed this is messy and is incomplete as it stands now.
Maybe this is not what OP wants? If so, I shall delete it then. And thanks for the attention.