Prove that no $n,m, 0Prove or disprove the claim that there are integers $n,m, 0<n<m$ such that $m^2 +mn+n^2$
is a perfect square.
 A: For instance, let $m=5$ and $n=3$. 
Remark: Here is a beginning to a general analysis. We want  $x^2+xy+y^2$ to be a perfect square. It is easier to multiply by $4$, and solve $4(x^2+xy+y^2)=z^2$.
Complete the square. So we want $(2x+y)^2+3y^2=z^2$. More generally look for solutions of $u^2+3v^2=z^2$. To produce solutions, we take a reasonable $v^2$, and express $3v^2$ as a difference of squares. 
Added: In a comment OP asked whether there are infinitely many (relatively prime) pairs $(m,n)$.
Let $n$ be odd. Note that 
$$\left(\frac{3n^2-1}{2}\right)^2 + 3n^2=\left(\frac{3n^2+1}{2}\right)^2.$$
So if $2m+n=\frac{3n^2-1}{2}$, then the pair $(m,n)$ should do the job. That gives $m=\frac{3n^2-2n-1}{4}$. (It is easy to check that $3n^2-2n-1$ is divisible by $4$). 
This by no means produces all pairs $(m,n)$ with the desired property.   
A: Hint $\rm\ \ A^2 + AB+ B^2 = (a^2+ab+b^2)^2\ $ for $\rm\ A = a^2-b^2,\,\ B = b^2 + 2ab$
This arises from the norm-multiplicativity $\rm\ N(\alpha^2) = (N(\alpha))^2\ $ for $\alpha$ an Eisenstein integer.
