Can we prove the existence of a gcd in $\mathbb Z$ without using division with remainder?

One can eliminate explicit use of the division algorithm by replacing its use by its inductive essence (i.e. rewriting the proof in mathematical "assembly language"). Below is one simple way.

The set $\rm\:S\:$ of all integers of the form $\rm\:a\:x + b\:y,\,\ x,y\in \mathbb Z,\:$ is closed under subtraction so, by the Lemma below, every positive $\rm\:n\in S\:$ is divisible by $\rm\:d = $ least positive $\rm\in S,\:$ so $\rm\:a,b\in S\:$ $\Rightarrow$ $\rm\:d\:|\:a,b.\:$ So $\rm\:d\:$ is a common divisor of $\rm\:a,b,\:$ necessarily greatest: $\rm\:c\:|\:a,b\:$ $\Rightarrow$ $\rm\:c\:|\:d = a\,x_1\!+\! b\,y_1\,$ ($\Rightarrow$ $\rm\:c\le d).$

Lemma $\ \ $ Let $\,\rm S\ne\emptyset \,$ be a set of integers $>0$ closed under subtraction $> 0,\,$ i.e. for all $\rm\,n,m\in S, \,$ $\rm\ n > m\ \Rightarrow\ n-m\, \in\, S.\,$ Then every element of $\rm\,S\,$ is a multiple of the least element $\rm\:\ell = \min\, S.$

Proof $\ $ If not there is a least nonmultiple $\rm\,n\in S,\,$ contra $\rm\,n-\ell \in S\,$ is a nonmultiple of $\rm\,\ell.$

Remark $\ $ Using the same idea one can give "elementary" proofs of other theorems of number theory, e.g. the proof of FTA = Fundamental Theorem of Arithmetic due to Zermelo (and others).

However, this is exactly the opposite of what one desires from a pedagogical standpoint. The innate ideal theoretic structure that underlies many of these results should be highlighted - not eliminated (for example, see posts on denominator ideals and order ideals)