That equation has no integer solutions. Remember that $\displaystyle x^2$ is congruent to 0 (mod 4) if $\displaystyle x$ is even, and congruent t0 1 (mod 4) if $\displaystyle x$ is odd.
So $\displaystyle a^2b^2\equiv0\pmod4$ unless $\displaystyle a$ and $\displaystyle b$ are both odd. But if $\displaystyle a$ and $\displaystyle b$ are both odd the $\displaystyle a^2+b^2+c^2$ will be congruent to 2 or 3 (mod 4), so that case cannot arise.
Therefore $\displaystyle a^2b^2\equiv0\pmod4$ and hence $\displaystyle a^2+b^2+c^2\equiv0\pmod4$. But that implies that $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$ are all even, say $\displaystyle a=2a'$, $\displaystyle b=2b'$, $\displaystyle c=2c'$. Then the equation becomes $\displaystyle a'^{\,2}+b'^{\,2}+c'^2=4a'^{\,2}b'^{\,2}$. But this time, the right side is obviously congruent to 0 (mod 4), so $\displaystyle a'$, $\displaystyle b'$ and $\displaystyle c'$ must all be even.
That argument can be continued indefinitely, so by
Fermat's method of infinite descent it follows that there are no solutions.