Solving the Diophantine equation $k^2(k+1)=m(3m-1)$

Question

I am working on solving the following Diophantine equation:

$$k^2(k+1)=m(3m-1)$$

And so far I solved, using WolframAlpha, the following solutions:

$$(k,m)=(-1,0);(0,0);(1,1);(4,-5);(6,-9)$$

Is there a way to prove that those are the only ones?

Answer 1

Hint: This is an elliptic curve given by $3y^2-y=x^3+x^2$. Its integer points can be computed, see for example the following post (we might transform to short Weierstrass form $y^2=x^3+ax+b$ if necessary).

How to compute rational or integer points on elliptic curves

Answer 2

(k,m)=(-2/3),(-1/9) & (k,m)=(-2/3),(4/9)

The above are also the solutions to the equation asked by "OP".