$x^2+y^2+xy=(x+y+\sqrt{xy})(x+y-\sqrt{xy})\implies \frac{133}{19}=7=x+y-\sqrt{xy}$ $(x+y+\sqrt{xy})+(x+y-\sqrt{xy})=19+7\implies x+y=13\implies \sqrt{xy}=6\implies xy=36$ By Vieta’s formula, $x,y$ are roots of $$a^2-13a+36=0\implies (a-4)(a-9)=0\implies a=4,9\implies \boxed{(x,y)=(4,9),(9,4)}$$

Set $m=x+y$ and $n^2=xy$. Then, $x^2+y^2+2xy = m^2\implies x^2+xy+y^2 = m^2-n^2$. Hence, $m^2-n^2=133$ and $m+n=19$. From here, $m-n=7$. Thus, $x+y=13$ and $xy=36$. Setting $y=36/x$ we obtain $x^2-13x+36=0\iff x\in\{4,9\}$, yielding $(x,y)=(4,9)$ and $(9,4)$.

Let $a=x+y$ and $b=\sqrt{xy}$. The first equation becomes $a^2-b^2=133$ and the second becomes $a+b=19$. Dividing we get $a-b=7$. From this we get $(a,b)=(13,6)$. So $(x,y)$ are the roots of the quadratic $z^2-13z+36=0$. From this we get $(x,y)=(4,9)$ or $(9,4)$.