The Diophantine equation $x^3+3=2^n$ has the obvious solutions $(-1,1)$,$(1,2)$ and $(5,7)$. I have been wondering if there are any other, but my attempts have been fruitless (I tried factoring it over $\Bbb{Q}(\sqrt[3]{3})$, but I don't know the basic properties of this field, such as what the ring of integers are, the class number etc.). Any help in solving this problem would be greatly appreciated.
Edit: We can actually split this into two, more general equations, namely the elliptic curves $$ x^3+3=y^2 \quad \text{and} \quad x^3+3=2y^2 $$ so this opens up another method for solving it.The first one is, in fact, a special case of the infamous Mordell equation.
Edit 2: Looking at this paper by Tzanakis and De Weger :http://www.math.uoc.gr/~tzanakis/Papers/PracticalSolutionThueEq.pdf I was wondering if we could use the methods explained in Section 3 and extend them to these equations (the methods in the paper, however, require some computational machinery).This could lead to solving a Thue equation, which has only finitely many integral solutions and the general method for solving them is in the paper linked.
Two cases:
$1)$ $n=2k$. Then $x^3+3=\left(2^k\right)^2$. But $a^3+3=b^2$ has $2$ integral solutions (http://oeis.org/A081119) $(a,b)=(1,\pm 2)$, so $(x,n)=(1,2)$.
$2)$ $n=2k+1$. Then $(2x)^3+24=\left(2^{k+2}\right)^2$. But $a^3+24=b^2$ has $8$ integral solutions (http://oeis.org/A081119, in particular http://oeis.org/A081119/b081119.txt)(I found them with a program, or you can see these tables)
$$(a,b)=(-2,\pm 4),(1,\pm 5),(10,\pm32),(8158,\pm736844),$$ so $(x,n)=(-1,1),(5,7)$.
Mordell Equations $x^3+k=y^2$ for $k\in\Bbb Z_{\neq 0}$ are fully solved when $|k|<10^4$ (were solved in $1998$; see here), so we can always use this method when the numbers aren't very large.