we have $\frac{a^2+b^2-c^2}{a+b-c}=a+b+c-2\frac{ab}{a+b-c}$ is an integer so $\frac{2ab}{a+b-c}$ is an integer and cyclicly.. hence $\frac{8(abc)^2}{(a+b-c)(a+c-b)(b+c-a)}=k$ or $16S^2=(a+b+c)(a+b-c)(a+c-b)(b+c-a)$ where $S$ is the area of the triangle hence $2kS^2=(a+b+c)(abc)^2$ let $p |gcd(k,a+b+c)$ a prime number where $p>2$ so $p | abc$ wolg $p |a$ and $p |a+b+c$ so $ p |b+c$ so $p |b+c-a$ so $p |bc$ hence $p$ divide $a,b,c$ contradiction if $gcd(k,a+b+c)=1$ we get $2k$ is a perfect square so $(a+b-c)(a+c-b)(b+c-a)$ is a perfect square. if $gcd(k,a+b+c)=1$ then $k$ is an odd perfect square so $2(a+b-c)(a+c-b)(b+c-a)$ is a perfect square.

@Muradjl 1) $16S^2$ not necessary perfect square for arbitrary $a,b,c.$ $\implies$ conditions $2kS^2=(a+b+c)(abc)^2$ and $gcd(k,a+b+c)=2^l$ don't give condition $k$. 2) You proved that if $p|gcd(k,a+b+c)\implies p = 2$ , but you only write about the case $gcd(k,a+b+c)=1$.

I will attempt to provide a more clear solution: Let $p$ be a prime larger than $2$ such that $p \mid (a+b-c)(b+c-a)(c+a-b)$. It suffices to prove that $\upsilon_p[(a+b-c)(b+c-a)(c+a-b)]$ is a multiple of $2$. Note that if $p$ divides at least two of the $a$, $b$, $c$, then it has to divide the third one and the condition $\gcd(a,b,c)=1$ is violated. We have: $$a+b-c \mid a^2+b^2-c^2 \Longleftrightarrow$$$$a+b-c \mid a^2+b^2-(a+b)^2 \Longleftrightarrow$$$$a+b-c \mid 2ab$$So, $\frac{2ab}{a+b-c}=x \in \mathbb{N}$. Similarly, $\frac{2bc}{b+c-a}=y \in \mathbb{N}$, $\frac{2ca}{c+a-b}=z \in \mathbb{N}$. Multiplying the equations, we get: $$xyz(a+b-c)(b+c-a)(c+a-b)=8a^2b^2c^2$$If $\gcd(p,xyz)=1$, the desired result is obvious. Otherwise, WLOG let $p \mid x$ and $p \mid a$. Hence, $p \nmid b$ and $p \nmid c$. In addition, $p \nmid y$ and $p \nmid b+c-a \Longrightarrow p \nmid b+c$. The initial conditions in light of $p \mid a$ gives: $$p \mid (b-c)^2(b+c) \Longleftrightarrow$$$$p \mid b-c$$Therefore, $p \mid a+b-c$ and $p \mid a+c-b$. If $\upsilon_p[a] \neq \upsilon_p[b-c]$, then $\upsilon_p[a+b-c]=\upsilon_p[c+a-b]=\min\{\upsilon_p[a],\upsilon_p[b-c]\}$, so $\upsilon_p[(a+b-c)(b+c-a)(c+a-b)]=2 \cdot \min\{\upsilon_p[a],\upsilon_p[b-c]\}$. If $\upsilon_p[a]=\upsilon_p[b-c]$, then $\upsilon_p[a+b-c] \geq \upsilon_p[a] \Longleftrightarrow \upsilon_p[\frac{a}{a+b-c}]=\upsilon_p[x] \leq 0$, which contradicts the fact $p \mid x$. Thus, in every possible case, $\upsilon_p[(a+b-c)(b+c-a)(c+a-b)]$ is a multiple of $2$ and the proof is complete.