Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

Let the gcd of $M$ and $N$ be $d$.

Let $M=dm$ and $N=dn$.
As $d$ is gcd, $gcd(m,n)=1$

So as given $lcm(M,N)=M^2-N^2+MN$

$lcm(M,N)=d^2m^2-d^2n^2+d^2mn=d^2(m^2-n^2+mn)$

We know, $gcd(a,b) \times lcm(a,b)=ab$

Using this,

$lcm(M,N) \times gcd(M,N)=MN$

$d^2(m^2-n^2+mn) \times d = dm \times dn$

$d^3(m^2-n^2+mn)=d^2mn$

$d(m^2-n^2+mn)=mn$

$dm^2-dn^2+dmn=mn$

$mn(d-1)=d(n^2-m^2)=d(n-m)(n+m)$

Note that $d$ is a divisor of Left hand side

As $gcd(d,d-1)=1$ and $d | mn(d-1)$ we have $d | mn$

As $gcd(m,n)=1$ 

So, $gcd(m,n-m)=1$

Also $gcd(m,n+m)=1$

So, $gcd(m,(n-m)(n+m))=1$

Similarly $gcd(n,(n-m)(n+m))=1$

So, $gcd(mn,(n-m)(n+m))=1$

which implies $mn | d$

We have $d | mn$ and $mn | d$ so $d=mn$

Now, $MN=dm \times dn =d^2mn=d^3$ which is a cubic number.