Suppose, $m,k,c$ are positive integers
Conjecture : The expression $$m^2k^2(c^2+1)^2-4mc(c^2-c+1)$$ is a perfect square if and only if $m=k=1$
In the case $m=k=1$ , we get $(c-1)^4$ which is a perfect square ($0$ and $1$ are allowed)
The hard part is to show that otherwise the expression cannot be a perfect square. I tried to compare $(mk(c^2+1)\pm 1)^2$ with the given expression but this led to nowhere. The conjecture is true for $m,k,c\le 1\ 600$
I arrived at this problem by trying to prove that for positive integers $a,b,c$ with $c^2+1\mid a+b$ and $ab\mid c(c^2-c+1)$ we have $a=c$ , $b=c^2-c+1$ or vice versa.
If we write $d=\gcd(m,4c(c^2-c+1))$, then there necessarily exist positive integers $a_1$ and $a_2$ such that $m=da_1^2$ and $$ mk^2(c^2+1)^2-4c(c^2-c+1) = d a_2^2. $$ Writing $t=a_1k$, we thus have $$ t^2 (c^2+1)^2 -a_2^2 = \frac{4c(c^2-c+1)}{d} $$ and so, taking $a_3=tda_2$ and multiplying by $t^2d^2$, $$ \left( t^2 d (c^2+1) \right)^2 - a_3^2 = 4 t^2dc(c^2-c+1). $$ If we define $a_4=t^2dc^2-2c+t^2d$, then $$ \left( t^2 d (c^2+1) \right)^2 - a_4^2 = 4 t^2dc(c^2+1)-4c^2. $$ If $t=d=1$, we are led to $a_3=a_4$ and so to the case $m=k=1$. Otherwise, $t^2d>1$ and hence necessarily have that $a_3 > a_4$. Checking that $a_3$ and $a_4$ have the same parity, it follows that $a_3 \geq a_4+2$. But $$ \left( t^2 d (c^2+1) \right)^2 - (a_4+2)^2 = 4 t^2dc(c^2-c+1) - (4t^2d + 4(c-1)^2) < 4 t^2dc(c^2-c+1). $$ This completes the proof. This argument is essentially Runge's method in disguise.