We have $$(19b+1)^2=a^8-a^7+1=(a^2-a+1)(a^6-a^4-a^3+a+1)$$Also $gcd(a^2-a+1,a^6-a^4-a^3+a+1)=1,\; 19$. But, $19$ doesn’t divide $(19b+1)^2$. Hence, the gcd should be $1$. This means both of the numbers are perfect squares. Then, $(2a-1)^2+3$ is a perfect square. Hence, $a=0,1$ and for these values $b=1$.

$x(x+2)$ reminds of $(x+1)^2 - 1$ so we have $a^8 - a^7 + 1 = (19b+1)^2$. from here we need to rewrite $a^8 - a^7 + 1 = xy$ such that $\gcd{(x,y)} = 1$. we know $a^8 - a^7 + 1 = (a^2-a+1)(a^6 - a^4 - a^3 + a + 1)$ and also with Euclidean algorithm we have $\gcd{(a^2-a+1,a^6 - a^4 - a^3 + a + 1)} = {1,19}$. Note that $(19b+1)^2 = 1$ mod $(19)$ so $\gcd{(a^2-a+1,a^6 - a^4 - a^3 + a + 1)} = 1$ so both $a^2-a+1,a^6 - a^4 - a^3 + a + 1$ are squares. we have $(a-1)^2 \le a^2 - a + 1 \le a^2$ so only cases which $a^2 - a + 1$ is square is when equalities hold so $a = {0,1}$, either one we have $(19b+1)^2 = 1 \implies b = 0$. Answers : $(a,b) = (0,0)$.