I was worried that my proof isn't right so I want to know if there are any mistakes in this and if this way can work? Thank you very much.
We want to show every maximal ideal $\mathfrak m$ of $\mathbb Z[\varphi]$ is principal. My idea is that the quotient $\mathbb Z[\varphi]/\mathfrak m$ must be a finite field, so we can use the classification of finite fields to produce a principal ideal equal to $\mathfrak m$.
Some lemmas which are used.
Lemma $\mathbb Z[\varphi]/\mathfrak m$ is isomorphic to either $\mathbb F_p$ (case 1) or $\mathbb F_{p^2}$ (case 2) (some $p$). proof: The quotient is a field since the ideal is maximal and the field must be finite (and have degree $\le 2$) because it's the quotient of a finite dimensional space ($\mathbb Z[\varphi]$ has dimension $2$) and doesn't contain $\mathbb Q$.
Edit, the next lemma is wrong but I don't use it anymore.
Lemma Suppose $\mathbb Z[\varphi]/\mathfrak m \simeq \mathbb Z[\varphi]/\mathfrak m'$ is an isomorphism then $\mathfrak m = \mathfrak m'$. proof: In general $R/I \simeq R/J$ implies $J$ is an automorphism of $R$ applied to $J$ but $\mathbb Z[\varphi]$ has no automorphisms.
Lemma $p$ is of the form $x^2 + xy - y^2$ iff $X^2 - X - 1$ has a solution mod $p$. proof: I don't actually prove this but I assume it could be done like this.
Note, if $p$ is of that form it can be written that way for infinitely many $(x,y)$ but $x+\varphi y$ will be an associate of $x'+\varphi y'$ or $x'+\bar\varphi y'$ if so.
The main theorem:
Lemma In case 1, $X^2 - X - 1$ has a solution $\mod p$. proof: The quotient map is a ring homomorphism.
Lemma If $X^2 - X - 1$ has a solution $\mod p$ then case 2 is impossible. proof: The basis $\{1,\varphi\}$ collapses as the image of $\varphi$ is expressible as an integer multiple of $1$, so $M$ is one dimensional.
Theorem $\mathfrak m$ is principal. proof:
In case 2, $\mathbb Z[\varphi]/(p)$ has size $p^2$ and it is the only ideal which gives a finite field of this size so $\mathfrak m = (p)$.
In case 1, Let $x$,$y$ such that $p = x^2 + xy - y^2 = (x + \varphi y)(x + \bar \varphi y)$ and both quotients like $\mathbb Z[\varphi]/(x + \varphi y)$ have size $p$ and there's no other way to quotient to get a field of the right size so either $\mathfrak m = (x + \varphi y)$ or $\mathfrak m = (x + \bar \varphi y)$ for any such associate pair $x$,$y$ (it doesn't matter which you choose).