I want to show that 7 is irreducible in $\mathbb Z[\sqrt{15}]$. I define the field norm $N(a+b\sqrt{15}) = a^2-15b^2$. If 7 can be factored into non-units, $7=\alpha\beta$, then $N(7)=49=N(\alpha)N(\beta)$. If $N(\alpha)=1,49$ then $\alpha$ or $\beta$ is a unit so I only want to show that $N(\alpha)\ne 7$. So suppose for contradiction that $N(\alpha)=7$. If $\alpha=a+b\sqrt{15}$ then $N(\alpha)=a^2-15b^2=7$.
From here I'm not sure how to solve it. This now seems to require some kind of number theory, yes? Having almost no background in number theory, this makes me wonder if there is a better way to approach the problem or some general method for showing that something like this has no solutions.
But if I try to think of some solution in number theory, then I guess computing mod 15 is natural.
$$ a^2 \equiv_{15} 7 $$
I can then try to think of the square of every number from 0 to 14, but this seems like a bad solution method.
There is nothing wrong with your solution method, you can work out all of the squares modulo $15$ and conclude that no solution can occur. Alternatively, you could have reduced modulo $5$ (less computation required), the squares modulo $5$ are just $\{\pm 1\}$, so as $7\equiv 2\mod 5$ it is not square mod $5$. In particular, no solution to $a^2-16b^2=7$ exists for $a,b$ integers.