Purpose of the first step in Engle-Granger cointegration test
...is to find a stationary combination of the integrated variables at hand. If there are just two variables, then it will be the stationary combination as it must be unique.
Why test $u_t$ (see below) for stationarity instead of any other linear combination?
Because when the stationary combination is unique, any other linear combination will be nonstationary regardless of whether the integrated variables at hand are cointegrated or not.
And to confirm, if $u_t$ is not stationary does it mean that $x_t$ and $y_t$ are not cointegrated?
Yes. If $u_t$ obtained from the first step of the Engle-Granger procedure is nonstationary, then $x_t$ and $y_t$ are not cointegrated.
Why won't I just test the following instead? $y_t-x_t=z_t$
Because that is very restrictive and need not hold in presence of cointegration. If two series $y_t$ and $x_t$ are cointegrated, there exists a $\beta$ such that $y_t-\beta x_t=u_t$ where $u_t$ is stationary. But there is no requirement that $\beta=1$.
Is $u_t$ somehow superior to $z_t$? In what way?
See the argumentation above. But when could $z_t$ be more useful than $u_t$? Say we have the following subject-matter knowledge of $\beta$: if cointegration is present (which we are not sure about and want to test for), $\beta=1$. Then we can just take $\beta=1$ which implies using $z_t$ in place of $u_t$ and test $z_t$ for a unit root. If, on the other hand, we ignored the knowledge and used $u_t$ rather than $z_t$, we could by chance end up finding a fit that is too good such that while $y_t$ and $x_t$ are not cointegrated in reality, $u_t$ still appears stationary (due to a small sample).