let $m,n \in N$ such that $2m^2+m=3n^2+n$, Find all integral solution of $2m^2+m=3n^2+n$? 
let $m,n \in N$ such that $2m^2+m=3n^2+n$. Find all integral solution of $2m^2+m=3n^2+n$ ?

My work:
$$2m^2-2n^2+m-n=n^2$$
$$\implies 2(m-n)(m+n)+(m-n)=n^2$$
$$\implies(2m+2n+1)(m-n)=n^2$$
Case $1$
$2m+2n+1=n^2$ and $m-n=1$
Now putting $m=n+1$ in $2m+2n+1=n^2$
which lead me to $2(n+1)+2n+1=n^2$
$$\implies n^2-4n-3=0$$
above equation give me irrational roots.
Case 2
$2m+2n+1=n$ and $m-n=n$
Now putting $m=2n$ in  $2m+2n+1=n$
which leads me to $2(2n)+2n+1=n$
$\implies n=\frac{-1}{5}$
which is not a natural number.
I am not getting any solution to the given equations. Is my approach correct?
Also is it necessary that $(m-n)$ and $(2m+2n+1)$ must be prefect square
Please Suggest a solution without using Modular Arithmetic
 A: $$\begin{align*}
2m^2+m&=3n^2+n \\
\frac{1}{8}\left((4m+1)^2-1\right)&=\frac{1}{12}\left((6n+1)^2-1\right)\\
3(4m+1)^2-3&=2(6n+1)^2-2
\end{align*}$$
Let $x=(4m+1)^2$ and $y=(6n+1)^2$, we have,
$$3x^2-2y^2=1$$
The above equation has been solved here
The general solution is obtained from the recurrences,
$$x_{n+1}=5x_n+4y_n,\;y_{n+1}=6x_n+5y_n,\;(x_1,y_1)=(1,1). $$
A: one may resolve the combined recurrence into separate degree two linear recurrences
$$  m_{k+2} = 98 \; m_{k+1}  - m_k + 24  $$
with the sequence beginning $0, 22, 2180, 213642, ... $
$$  n_{k+2} = 98 \; n_{k+1}  - n_k + 16  $$
with the sequence beginning $0, 18, 1780,  174438, ... $
The common values $v= 2 m^2 + m = 3 n^2 + n$ begin
$$  0, 990, 9506980, 91286021970, 876528373449960, 8416425350580494950, $$
and obey
$$  v_{k+2} = 9602 \; v_{k+1}  - v_k + 1000  $$
There was a question about $d_k = m_k - n_k.$  This is a square, $d_k$  beginning $0, 4, 400,...$ and
$$  d_{k+2} = 98 \; d_{k+1}  - d_k + 8  $$
If we name $s_k = \sqrt{m_k - n_k}$ we find that $s_k$ are integers, beginning $0, 2, 20, 198,...$  and obeying
$$  s_{k+2} = 10 \; s_{k+1}  - s_k   $$
