Determine all pairs of positive integers $(a,b)$ such that $a^2+b^2+ab$ is a perfect square. Consideration via mod $4$ shows that $a,b=0$ mod $4$ or one of between $a,b$ is $=1$ mod $4$ while the other is $=0 $ mod $4$.
Considering $(a+b)^2,a^2+an+b^2,(a+b-1)^2$ as we can deduce that $a,b>2$.

Characterization of all solutions (non-trivial) to the Diophantine equation $$a^2+ab+b^2=x^2\quad a,b,x\in\mathbb{Z}$$ using Algebraic Number Theory

$\mathrm{WLOG}$ assume that $\gcd(a,b)=1$.We know that $\mathbb{Z}[\omega]$ is a $\mathrm{UFD}$ where $\omega=\frac{1+\sqrt{-3}}{2}$. Now we can factorize $$a^2+ab+b^2=(a-b\omega)(a-b\omega^2)=(a-b\omega)(a+b+b\omega)$$ in $\mathbb{Z}[\omega]$. Let $a^2+ab+b^2=x^2$ for some $x\in\mathbb{N}$. If $\pi\mid(a-b\omega)$ and $\pi\mid(a+b+b\omega)$ then we can show that $\pi\mid1$ and hence $\gcd(a-b\omega,a+b+b\omega)=1$. Then by unique factorization property we conclude $\exists$ $\alpha-\beta\omega,\gamma-\delta\omega\in\mathbb{Z}[\omega]$ such that $a-b\omega=u(\alpha-\beta\omega)^2=u(\alpha^2-2\alpha\beta\omega-\beta^2(1+\omega))=u((\alpha^2-\beta^2)-(2\alpha\beta+\beta^2)\omega)$ and $a+b+b\omega=v(\gamma-\delta\omega)^2$ where $u,v$ are units and hence $u,v\in\{\pm1,\pm\omega,\pm\omega^2\}$. We consider the case $u=1$ only, the other cases gives similar solutions. Comparing we get $a=\alpha^2-\beta^2$ and $b=\beta^2+2\alpha\beta$. We see that in fact $\gamma-\delta\omega=\overline{\alpha-\beta\omega}=\alpha-\beta\omega^2$ and hence $(a^2+ab+b^2)=(\alpha^2+\alpha\beta+\beta^2)^2$. So the set $$\{(\alpha^2-\beta^2,\beta^2+2\alpha\beta,\alpha^2+\alpha\beta+\beta^2):\alpha,\beta\in\mathbb{Z},\gcd(\alpha,\beta)=1\}$$ describes all solutions $(a,b,x)$ to the Diophantine equation $$a^2+ab+b^2=x^2$$ Also we observe that $(a,b,x)$ is a solution if and only if $(a,b,-x),(-a,-b,-x),(-a,-b,x),(b,a,x),(b,a,-x),(-b,-a,-x),(-b,-a,x)$ are solutions. These solutions will be obtained by considering other units except $u=1$. These observations are trivial.

