The form of the reduced matrix tells you that everything can be expressed in terms of the free parameters $x_3$ and $x_4$. It may be helpful to take your reduction one more step and get to $$\pmatrix{4&0&1&2\cr0&4&3&2\cr}$$ Now writing $x_3=s$ and $x_4=t$ the first row says $x_1=(1/4)(-s-2t)$ and the second row says $x_2=(1/4)(-3s-2t)$. If we don't like fractions, we can instead write $x_3=-4u$, $x_4=-4v$, whence $x_1=u+2v$, $x_2=3u+2v$. So we have $$(x_1,x_2,x_3,x_4)=(u+2v,3u+2v,-4u,-4v)=u(1,3,-4,0)+v(2,2,0,-4)$$ Notice this is also equivalent to $span\{{(1,3,-4,0),(2,2,0,-4)\}}$ due to the parameters $u$ and $v$. Also notice that these two vectors are linearly independent.

Now you can read off the basis, $\{{\,(1,3,-4,0),(2,2,0,-4)\,\}}$.

This set is a basis because a) It is linearly independent, and b) because it spans the solution space