The difference between the two equations is

$$(a-b)(a^2+ab+b^2+6a+6b)=1863$$

$1863$ factors as $3^4\cdot23$ and its divisors are $1,3,9,23,27,69,81,207,621,1863$ so $a-b$ is one of $1,3,9,23,27$. Using this we can solve for $a$ in terms of $b$ and write the quadratic in our original factorization purely in terms of $b$. If it has a positive integer root for $b$ then that root is a solution. Applying this we find the quadratic has a positive integer root for $a=b+3$, specifically $b=11$.

All $(a,b)$ of positive integers such that $a^3âˆ’6b^2=2018$ and $b^3âˆ’6a^2=155$ hold simultaneously are $(14,11)$.