How do I find the generator $(g)$ that generates $(29, \sqrt{-5} ± 13)$ 
Find the generator $(g)$ that generates $(29, \sqrt{-5} + 13)$. The ring is $\mathbb{Z}\left[\sqrt{-5}\right]$.

The fact I used was that $\text{Norm}(g)$ must divide both 29 and $\text{Norm}(\sqrt{5} + 13)$, which means solving the Pell equation $x^2 + 5y^2 = 29$. The solutions to this Pell equation are $(\pm 3, \pm 2)$. I don't know what to do with the $\pm$ signs so let's just refer to $(3, 2)$ for simplicity.
Because ideals are subsets, proving ideal equality is proving two subset inclusions.

Show that $(29, \sqrt{-5} + 13) \subseteq (3 + 2\sqrt{-5})$ and that $(3 + 2\sqrt{-5}) \subseteq (29, \sqrt{-5} ± 13)$. (Because I do not know what to do with the signage, this may not be true.

How do I continue? 

To show that $(29, 13+\sqrt{-5}) \subseteq (g)$ we have to find integers $x,y,z,t$ such that $g(x + y\sqrt{-5})=29$ and $g(z + t\sqrt{-5})=13+\sqrt{-5}$. 
If $g=3 + 2\sqrt{-5}$ then 
$\cases{3z-10t=13\\ 
2z+3t=1}.$
That is $z=\tfrac{49}{29}$ and $t=-\tfrac {23}{29}$, which is impossible. 
Let's try $g=3 - 2\sqrt{-5}$ (the other sign choices are reduced to considered by multiplication of $g$ by $-1$). Then  
$\cases{3x+10y=29\\ 
-2x+3y=0}$
$\cases{3z+10t=13\\ 
-2z+3t=1}$
That is $x=3$, $y=2$, $z=1$, and $t=1$.
To show that $(3 - 2\sqrt{-5}) \subseteq (29, 13+\sqrt{-5})$ we have to find integers $x,y,z,t$ such that $$(x + y\sqrt{-5})29+(13+\sqrt{-5})(z + t\sqrt{-5})=3-2\sqrt{-5}.$$ That is 
$\cases{29x+13z-5t=3\\ 
29y+z+13t=-2}$
$-13z=29x-5t-3=13(29y+13t+2).$
If $y=0$ then $29(x-1)=174t$ that is $x-1=6t$. Thus we can put $x=1$, $y=0$, $z=-2$, and $t=0$.
