Consider the following equation$$a^3+300^2=k(a^2b^2+300),$$where $a$, $b$ and $k$ are positive integers.
Notice that $a^2b^2+300|a^3+300^2$ and $$a^2b^2+300|b^2(a^3+300^2)-a(a^2b^2+300)=300(300b^2-a)$$Now we have three cases:
$\color{red}{Case \ 1}$ : $300b^2-a<0$ then $$300(a-300b^2) \ge a^2b^2+300 \Longrightarrow b^2 \le \frac{300(a-1)}{a^2+300^2}<\frac{600a}{a^2+300^2}\le 1,$$which is not possible.
$\color{red}{Case \ 2}$ : $300b^2-a=0$, which gives a family of solutions.
$\color{red}{Case \ 3}$ : $300b^2-a>0$ then $$300(300b^2-a) \ge a^2b^2+300 \Longrightarrow 300^2b^2-300a \ge a^2b^2+300,$$which is possible only when $a<300$. Also, it follows from original equation that $k<300$. Otherwise, RHS of original equation will be larger than the LHS, since $a<300$.
Now we have $a, k<300$. A computer search shows that solutions are$$(a, b, k)=(10, 2, 130) \\ (a, b, k)=(20, 1, 140) \\ (a, b, k)=(30, 2, 30)$$Solutions are very simple and are related to each other in many ways. Like $a|k$, $a \equiv k \equiv 0 \pmod{10}$. I do not know why I cannot find an analytic way to solve case 3. It seems that solution should be easy. What do you think?
Some other points: If you let $d=\gcd(a, k)$, then for co-prime integers $a'$ and $k'$ we have$$a'^3d^3+300^2=k' a'^2 b^2 d^3+ 300 k' d$$Thus $d|300^2$ and $d=2^x3^y5^z$ for some non-negative integers $x$, $y$ and $z$. I proved using $d^3 | 300(300-k'd)$ that $x\le 2$, $y \le 1$ and $z \le 2$.
Also note that $a^2 |300(300-k)$
I think there must be something easy that I cannot see. I was working on this for a very long time! Come and rescue me :))