Only certain values seem to make an expression into perfect square, can you help me prove or disprove it? All the variables in this question are in integers. I am trying to prove that
$$\frac{4k-\Delta^2}3$$
is a perfect square only if $\Delta \in \{\pm(2a+b),\pm(a+2b),\pm(a-b)\}$ where $a,b$ are such that $k=a^2+ab+b^2$.
Example: $k=49$. The above expression becomes perfect square only for $\Delta \in \{2,7,11,13,14\}$ and their negatives. 
$7$ and $14$ can be expressed as $a-b$ and $2a+b$ for $a=7,b=0$ which satisfy $7^2+7\cdot 0 + 0^2=49$.
$2, 11$ and $13$ can be expressed as $a-b, a+2b$ and $2a+b$ for $a=5, b=3$ which satisfy $5^2+5\cdot 3 + 3^2=49$.
Could you give me some hints on how to approach this and prove that any valid $\Delta$ will be expressable in those forms using $a,b$ that satisfy $k=a^2+ab+b^2$?
The background
To give some depth to my question - I was investigating the following equation in integers:
$$z^2=\frac{4k-\Delta^2}3$$
I discovered that it is solvable iff $k$ is a Löschian number (exists $a,b$ such that $k=a^2+ab+b^2$). 
The proof: 
For a Löschian $k=a^2+ab+b^2$ take any $\Delta \in \{\pm(2a+b),\pm(a+2b),\pm(a-b)\}$, insert $k,\Delta$ in RHS the and get a square. For the opposite direction - if $z, k, \Delta$ satisfy the equation, take $a=z$, $b=\frac{\Delta-z}2$ (can be shown to be integer) and see that $a^2+ab+b^2=\frac{3z^2+\Delta^2}4=k$ therefore $k$ is Löschian.
However, I also strongly suspect (based on numerical experiments) that the listed $\Delta$ values are the only ones that satisfy the equation. But I struggle to prove that.

Suppose $\frac{4k - \Delta^2}{3}$ is a square, in particular say $$\frac{4k - \Delta^2}{3}=M^2.$$ Then $4k = \Delta^2 + 3 M^2$. 
Reducing modulo 4, we see that $\Delta^2 + 3M^2 = 0 \pmod{4}$, so $\Delta^2 = M^2 \pmod{4}$. Hence, $\Delta=M \pmod{2}$, so they are both even or both odd.

If they are both odd:
Set $$a = \frac{\Delta + M}{2}, \quad b = \frac{M - \Delta}{2}$$ which are certainly both integers.
It can be checked that $a^2 + ab + b^2 = \frac{1}{4}(\Delta^2 + 3M^2) = k$. 
Also, $\Delta = a-b$, so $\Delta$ can be written as $a-b$, where $a,b$ satisfy $k=a^2 + ab+b^2$

If they are both even:
Similarly, set $$a = \frac{\Delta-M}{2}, \quad b=M$$ both integers. 
Again, we have $a^2 + ab+b^2 = k$, and $\Delta = 2a+b$, as required. 

Ultimately, we have shown that for any $\Delta$ which makes $\frac{4k-\Delta^2}{3}$ a square number, there are $a,b$ satisfying $a^2 + ab+b^2=k$ and $\Delta = 2a+b$ or $\Delta = a-b$.
