Conflict with the properties of convolution excuse me in advance for the lenghty question, consider the following signals:

$x_1[n]$, $x_2[n]$, $y[n]$ and

$z_1[n] = x_1[n] * y[n]$

$z_2[n] = x_2[n] * y[n]$

In addition, we know that:

$x_2[n] = x_1[n] - x_1[n-3]$

Thus, we can write:

$z_2[n] = (x_1[n] - x_1[n-3])*y[n]$

$z_2[n] = (x_1[n]*y[n]) - (x_1[n-3]*y[n])$

$z_2[n] = z_1[n] - (x_1[n-3]*y[n])$

Now for the question; the solution to the exercise I'm working on states that:

$z_2[n] = z_1[n] - z_1[n-3]$

Now this would mean that the following:

$z_1[n-3] = (x_1[n-3]*y[n])$ is true.

Isn't $z_1[n-3] = x_1[n-3]*y[n-3]$ ?

What makes me hesitate is that this development seems to support the given answer:

$x_1[n-3] = x_1[n]*\delta[n-3]$

$x_1[n-3] * y[n] = (x_1[n]*\delta[n-3])*y[n]$

$x_1[n-3] * y[n] = (x_1[n]*y[n])*\delta[n-3]$

$x_1[n-3] * y[n] = z_1[n]*\delta[n-3] = z_1[n-3]$

Which answer is correct and why is there a conflict with both reasonings? Thank you!
I believe that the problem you have with this exercise is mainly caused by bad notation. Writing convolution as

$$z[n]=x[n]\star y[n]\tag{1}$$

(which I think is bad notation) it is tempting to believe that

$$z[n-1]\stackrel{?}{=}x[n-1]\star y[n-1]\tag{2}$$

which is not the case.

If you take the definition of (discrete) convolution

$$z[n]=\sum_kx[k]y[n-k]\tag{3}$$

then it's obvious that

$$z[n-1]=\sum_kx[k]y[n-1-k]=\sum_kx[n-1-k]y[k]\tag{4}$$

i.e., only one of the two signals needs to be delayed by one sample, not both.

A better notation is

$$z[n]=(x\star y)[n]\tag{5}$$

Now you can define $x_1[n]=x[n-1]$ and $y_1[n]=y[n-1]$ and you have

$$z[n-1]=(x\star y_1)[n]=(x_1\star y)[n]\tag{5}$$

If you view $z[n]$ as the output of a system with impulse response $y[n]$ and input $x[n]$, then it should also be obvious that delaying $z[n]$ by one sample is equivalently achieved either by delaying the input signal, or by adding a delay to the impulse response, but not by doing both.