We know that for two signals $$x[n]$$ and $$h[n]$$ such that : $$y_{1}[n]=(x*h)[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k]=\sum_{k=-\infty}^{\infty}h[k]x[n-k]$$ We can deduce that :$$y_{2}[n]=x[n+2]*h[n]=y_{1}[n+2]$$ and $$y_{3}[n]=x[n]*h[n+2]=y_{1}[n+2]$$ but can we know what : $$y_{4}[n]=x[n+2]*h[n+2]$$ gives in terms of $$y_{1}[n]$$? I am unaware if I might be asking a really stupid question or am I not seeing it? Is it $$y_{1}[n+4]$$?

Whenever you have doubts regarding the properties of the convolution operator, you should resort to its definition.

Let $$x_s[n] = x[n+2]$$ and $$h_s[n] = h[n+2]$$. Then:

$$y_4[n] = x_s[n] * h_s[n] = \sum_{k=-\infty}^\infty x_s[k]h_s[n-k]$$

If we replace with the definitions of our discrete functions:

$$y_4[n] = \sum_{k=-\infty}^\infty x[k+2]h[n-k+2]$$

See what happens if we make the change of variables $$m = k+2$$:

$$y_4[n] = \sum_{m=-\infty}^\infty x[m]h[n-(m-2)+2]=\sum_{m=-\infty}^\infty x[m]h[n+4-m]$$

Does that final expression sound familiar? You were right indeed in your OP, as that is exactly what you thought:

$$y_4[n] = \sum_{m=-\infty}^\infty x[m]h[n+4-m] = y_1[n+4]$$