Is there a way of solving the following equation, in integers $(x,y)$, by hand? :
$2^{3x}+17=y^2$.
You can also try: $2^{2x}+17=y^2$ or more generally $2^x+17=y^2$; each of these has at least 1 solution.
I used the elliptic curve $x^3+17=y^2$ to make this, since it had 3 solutions where $x$ is a power of 2. These are:
$(x,y)\in \{(2,\pm5), (4,\pm9), (8,\pm23)\}$.
$2^x+17=y^2$ leads to the three elliptic curves $y^2=u^3+17$, $y^2=2u^3+17$, and $y^2=4u^3+17$ (depending on $x$ modulo $3$). There are standard techniques for solving these, that is, for finding all the integral solutions. Indeed, you can write each one as a Mordell equation, $Y^2=U^3+k$; for example, from $y^2=4u^3+17$ we get $16y^2=64u^3+272$, so $Y^2=U^3+272$. Solutions to Mordell equations for various ranges of values of $k$ have been tabulated, and it should be easy to find the tables by searching the web for "Mordell equation".
$2^{2x}+17=y^2$ is easy. It's $y^2-r^2=17$, which has only the solutions $y=\pm9$, $r=\pm8$, so $x=3$.