The answer is NO. Assume that it is possible. Let the row sums are $2^a, 2^b, 2^c, 2^d$ and column sums are $2^p, 2^q, 2^r, 2^s$ where $a,b,c,d,p,q,r,s$ are all different. Now, the total sum $= 2^a+2^b+2^c+2^d=2^p+2^q+2^r+2^s$ WLOG, assume that $a<b<c<d$ and $p<q<r<s$ $\implies 2^a(1+2^{b-a}+2^{c-a}+2^{d-a})=2^p(1+2^{q-p}+2^{r-p}+2^{s-p})$ Clearly both the expressions inside the brackets are odd as the eight numbers are different, hence by equating the order of 2 $\implies 2^a=2^b \implies a=b$, which contradicts our initial assumption. So, there is no such configuration.