Let $n^2+m=(n+c)^2$ where $c\ge 1$. Then, $m=2nc+c^2$. Hence, $2nc+c^2|3n^2$. Let $3n^2=2knc+c^2k$ where $k\ge 1$. We have $kc^2+2knc-3n^2=0$. Then, $\triangle_c=(2kn)^2+4(3n^2)k=4k^2n^2+12n^2k=4n^2(k^2+3k)$. This number needs to be a perfect square. Then, $k^2+3k$ is a perfect square. We have $(k+2)^2>k^2+3k\ge (k+1)^2\Rightarrow k^2+3k=(k+1)^2\Rightarrow k=1$. Then, $2nc+c^2=3n^2\Rightarrow m=3n^2$. Clearly, these numbers satisfy the conditions. Hence, $\boxed{\text{ALL SOLUTIONS:} (m,n)=(3t^2, t)}$ where $t\in\mathbb{Z^+}$.

Setting $n^2+m=(n+k)^2$ for some $k$, we have $m=2nk+k^2$. Let $d={\rm gcd}(k,n)$ with $k=dk_1$ and $n=dn_1$. We then have \[ m\mid 3n^2\implies d^2\bigl(2n_1k_1+k_1^2\bigr) \mid 3n^2 = 3d^2n_1^2\implies 2n_1k_1+k_1^2\mid 3n_1^2. \]Since $(k_1,n_1)=1$, it follows $k_1\in\{1,3\}$. For $k_1=1$, we find $2n_1+1\mid 3n_1^2$. Clearly, $n_1\equiv -\frac12\pmod{2n_1+1}$, and from here it follows $3n_1^2\equiv \frac34\pmod{2n_1+1}$, and consequently $2n_1+1\mid 3$. That is, $n_1=1$. With this, $n=k=d$ and $m=3d^2$. Namely, the parametric family $(3d^2,d)$ for any $d\in\mathbb{N}$ is a solution. A similar reasoning for $k_1=3$ implies $2n_1+3\mid 9$, that is $n_1=3$ which is not possible. Hence, $(3d^2,d),d\in\mathbb{N}$, is the only solution.

Let m=a^2-n^2 , and let gcd (a,n)=d , so a=dx ,n=dy where gcd(x,y)=1 3d^2.y^2/(x^2-y^2).d^2 , x^2-y^2 divides 3y^2 , and gcd(x^2-y^2 , y^2)=1 , x^2-y^2 divides 3 which makes (x,y)= (2,1) so m=3d^2 ,n=d. ANSWER= ( 3d^2, d)

CahitArf wrote: Let $m=a^2-n^2$ , and let $gcd (a,n)=d$ , so $a=dx ,n=dy$ where $gcd(x,y)=1$ $3d^2.y^2/(x^2-y^2).d^2$ , $ x^2-y^2$ divides $3y^2$ , and $gcd(x^2-y^2 , y^2)=1$, $x^2-y^2$ divides $3$ which makes $(x,y)$= $(2,1)$ so $m=3d^2$, $n=d$. ANSWER= $( 3d^2, d)$

Thanks a lot Jjesus . I don't know how to write in latex .

Let $\dfrac{3n^2}{m}=k\in\mathbb{N},$ then $\sqrt{n^2+\dfrac{3n^2}{k}}\in\mathbb{N}\implies\sqrt{1+\dfrac{3}{k}}\in\mathbb{Q}.$ $3\nmid k\implies$ (as $\gcd(k,k+3)=1$) $\sqrt{k+3}\in\mathbb{N},\sqrt{k}\in\mathbb{N}\implies k=1\implies m=3n^2.$ $k=3t,t\in\mathbb{N}\implies\sqrt{\dfrac{t+1}{t}}\in\mathbb{Q}\implies\sqrt{t}\in\mathbb{N},\sqrt{t+1}\in\mathbb{N},$ a contradiction. Finally $(m,n)=(3z^2,z)$ satisfy the equation for $z\in\mathbb{N}.$

We will look at two cases. $1)$ If $ 3|m $ $m=3x , x=a^2b $ where b is an square-free number. $m| 3n^2 \Rightarrow x| n^2 \Rightarrow a^2|n^2 \Rightarrow a|n \Rightarrow n=ac $ and because $x|n^2 \Rightarrow a^2b | a^2c^2\Rightarrow b|c^2 \Rightarrow b|c \Rightarrow n = abd, m= 3a^2b$ $n^2+m = a^2(b^2d^2+3b)$ and since$(bd)^2 < b^2d^2 + 3b<(bd+2)^2, b^2d^2+3b = b^2d^2+2bd+1 \Rightarrow b=d=1, \boxed{n=a, m= 3a^2}$ $2) 3 \nmid m$ Similarly $m=a^2b$ where b is a squarefree positive integer. $m| 3n^2 \Rightarrow m|n^2 \Rightarrow a^2|n^2 \Rightarrow a|n \Rightarrow n=ac $ and because $m|n^2 \Rightarrow a^2b | a^2c^2\Rightarrow b|c^2 \Rightarrow b|c \Rightarrow n = abd, m= a^2b$ $n^2+m = a^2(b^2d^2 +b)$, however $(bd)^2<b^2d^2+b<(bd+1)^2$ a contradiction. So the only solution is $\boxed{n=a, m= 3a^2}$ for any $a\in\mathbb{Z^+}$

Hello mates! $\sqrt{n^2+m}=x$ $\frac{3n^2}{m}=y$ $m=\frac{3n^2}{y}$ So $n \sqrt{\frac{3+y}{y}}=x$ So that 3n²=m And its solutions $n=t^2;m=3t^2$

$\color{blue} \boxed{\textbf{SOLUTION}}$ Let, $\frac{3n^2}{m}=l \implies 3n^2=ml, l \geq 1$ Now, $\sqrt{n^2+m}=n \sqrt{1+ \dfrac{3}{l}}$ So, $\sqrt{1+\dfrac{3}{l}} \in \mathbb{Z^+}$ If, $l > 1,$ $0 < \frac{3}{l} < \frac{3}{1}=3$ $\implies 1<1+ \dfrac{3}{l} <4$ $\implies \sqrt{1+\dfrac{3}{l}}$ not $\in \mathbb{Z^+}$ So, $l=1$ And we get, $\color{red} \boxed{\text{} (m,n)=(3t^2, t)}$

$\begin{cases} \frac{3n^2}{m} = x^2 \\ n^2 + m = y^2 \end{cases}$ $x, y \in \mathbb{Z_+^*}$ $\bullet gcd(m, 3) = 3$ $\Longrightarrow m = 3k \Longrightarrow$ $\frac{n^2}{k} = x^2 \Longrightarrow$ $n^2 = x^2k \Longrightarrow k = t^2$ $\Longrightarrow n^2 + m = x^2k + 3k = y^2$ $\Longrightarrow x^2 + 3 = (\frac{y}{t})^2$ $\Longrightarrow 3 = (\frac{y}{t} + x)(\frac{y}{t} - x)$ $\Longrightarrow$ $\begin{cases} \frac{y}{t} + x = 3 \\ \frac{y}{t} - x = 1 \end{cases}$ Then $x = 1, y = 2t \Longrightarrow x = 1$ $\Longrightarrow (m, n) = (3t^2, t)$ $\forall t \in \mathbb{Z_+^*}$ $\bullet gcd(m, 3) = 1$ Written $m = dm_0, n = dn_0$ with $gcd(m_0, n_0) = 1$ Thus $\frac{3n^2}{m} = \frac{3dn_0^2}{m_0} \Longrightarrow m_0 \mid d$ $\Longrightarrow d = m_0l \Longrightarrow$ $n^2 + m = d^2n_0^2 + dm_0 =$ $m_0^2n_0^2l^2 + m_0^2l = y^2$ $\Longrightarrow n_0^2l^2 + l =$ $(n_0^2l + 1)l = (\frac{y}{m_0})^2$ but $gcd(n_0^2l + 1, l) = 1$ so, $\begin{cases} n_0^2l + 1 = \alpha^2 \\ l = \beta^2 \end{cases}$ $\Longrightarrow (n_0\beta)^2 + 1 = \alpha^2 \Longrightarrow$ $\begin{cases} n_0 = 0 & \Longrightarrow n = 0 ABS! \\ \beta = 0 & \Longrightarrow l = 0 \Longrightarrow d = 0 ABS! \end{cases}$ Hence this case has no solution. Therefore, $(m, n) = (3t^2, t)$