I guess the $ (x,y)$ should be $ (a,b)$ or vice versa... By Fermat's little theorem, $ (19a + b)^{18},(a + b)^{18},(19b + a)^{18}\in\{0,1\}\pmod{19}$. Since $ 2$ and $ 3$ aren't squares modulo $ 19$, we need the sum it to be $ 0$ or $ 1$. If $ 19\not|\ a + b$ then we need $ 19|18a$ and $ 19|18b$ by the above, contradiction. So $ 19|a + b$. If $ 19\not|\ a$ and $ 19\not|\ b$ then the sum is $ 2$, contradiction. So at least one divides $ 19$, and by the above, the other does as well. So we need $ 19|a,b$. Now take a nonzero solution where $ |a| + |b|$ is minimal. By the above, $ \left(\frac {a}{19},\frac {b}{19}\right)$ is an integer solution, and it has $ \left|\frac {a}{19}\right| + \left|\frac {b}{19}\right| < |a| + |b|$, contradiction. So we only have the zero solution: $ a = b = 0$.

sketch: 1. if all a,b,c are not divisible by 19 ,then the term is equivalent to 3 mod 19 (by fermat little theorem) but 3 is not a quadratic residue modulo 19 (since $3^9 = -1 (mod 19)$ ) ,contradiction! 2.if one of a,b (wlog ,a) is divisible by 19 so the term is equivalent to 2 mod 19, again , 2 is not a quadratic residue modulo 19 contradiction! 3.let $a=19a_1 , b=19b_2$ then the term will be $19^{18}((19a_1+b_1)^{18}+(a_1+b_1)^{18}+(19b_1+a_1)^{18})$ but$ (19a_1+b_1)^{18}+(a_1+b_1)^{18}+(19b_1+a_1)^{18}$ is not perfect squares as proved above, contradiction. so, no solution. (a=0,b=0 gives 0 which is not satisfies the problem condition)