Back to 1957!

Let $x+y=a$ and $xy=b$ . Then $x^{3}+y^{3}=1957\\ \implies (x+y)^{3}-3xy(x+y)=a^{3}-3ab=1957 \cdots (1).$ $(x+y)(x+1)(y+1)=2014\\ \implies (x+y)(x+y+xy+1)=a(a+b+1)=a^{2}+ab+a=2014 \cdots (2).$ From $(1)$ , $ab=\frac{a^{3}-1957}{3}\cdots(3).$ Substituting $(3)$ into $(2)$ , $a^{2}+\frac{a^{3}-1957}{3}+a=2014\\ 3a^{2}+a^{3}-1957+3a=6042\\ a^{3}+3a^{2}+3a-7999=0\\ a^{3}+3a^{2}+3a+1=8000\\ (a+1)^{3}=20^{3} \implies a=x+y=20-1=\boxed{19}.$