$x,y$ are positive integers.
$$13^x+3=y^2\iff \left(4+\sqrt{3}\right)^x\left(4-\sqrt{3}\right)^x=\left(y+\sqrt3\right)\left(y-\sqrt3\right)$$
$\gcd\left(y+\sqrt3, y-\sqrt3\right)=1$, therefore $y+\sqrt3=\left(4+\sqrt{3}\right)^x$ (implies $x=1$), since $\Bbb Z\left[\sqrt{3}\right]$ is a UFD and $4+\sqrt{3}$ is prime (since $4^2-3=13$ is prime) and $y-\sqrt{3}$ has negative $\sqrt3$ part.
My question: why can't RHS have units, e.g. $y+\sqrt3=\left(2+\sqrt3\right)\left(4+\sqrt{3}\right)^x$.
(then $y-\sqrt3=\left(2-\sqrt3\right)\left(4-\sqrt{3}\right)^x$)
Solution found in Introduction to Diophatine Equations by Andreescu Titu (page $315$).