I'm puzzled by the following:
if $R$ is a unitary ring then $R$ is generated by $1_R$, denoted as $$R = \langle 1_R \rangle. $$ can we conclude that every unitary ring is finitely generated? I know the answer is no, as $\mathbb{Q}$ is not finitely generated, although we can express $\mathbb{Q}$ as $\langle 1 \rangle$. Can somebody help explain what's causing my confusion?
Your problem lies in the fact that you are confusing being finitely generated as an ideal and being finitely generated as a module.
Taking any ring $R$, it is true that $R=(1)$ as an ideal, which is equivalent to say that $R=\langle 1\rangle _R$ as an $R$-module.
So $\mathbb{Q}=\langle 1\rangle _\mathbb{Q}$ as a $\mathbb{Q}$-module (i.e. as an ideal and thus, according to your definition, as a ring).
However $\mathbb{Q}$ is also a $\mathbb{Z}$-module. If you consider $\mathbb{Q}$ with this structure, as you correctly pointed out, it is not finitely generated. But this means that it is not finitely generated as a $\mathbb{Z}$-module, not as an ideal ($=\mathbb{Q}$-module).