Which triangular numbers are also squares? I'm reading Stopple's A Primer of Analytic Number Theory:

Exercise 1.1.3: Which triangular numbers are also squares? That is, what conditions on $m$ and $n$ will guarantee that $t_n=s_m$? Show that if this happens, then we have:
$$(2n+1)^2-8m^2=1,$$
a solution to Pell's equation, which we will study in more detail in Chapter $11$.

I thought about the following:
$$\begin{eqnarray*}
  {t_n}&=&{s_n} \\ 
  {\frac{n^2+n}{2}}&=&{m^2} \\ 
  {n^2+n}&=&{m^2} 
\end{eqnarray*}$$
I've solved for $n$ and $m$ but I still have no clue of how to proceed. I've looked at the book's solution and the solution is as follows: 
$$\begin{eqnarray*}
  {\frac{n(n+1)}{2}}&=&{m^2} \\ 
  {n(n+1)}&=&{2m^2} \\ 
  {\color{red}{4n(n+1)}}&\color{red}{=}&{\color{red}{8m^2}} \\
  {4n^2+4n+1-1}&=&{8m^2} \\
  {(2n+1)^2-1}&=&{8m^2} \\
\end{eqnarray*}$$
In the red line, he multiplies the equation by $4$, I don't understand why to do it nor how the condition is achieved.

This conversion course gives solutions, but there are obtained two equations.
Certainly better to solve the equation.
Of course the solution of equation:    $Y^2=\frac{X(X\pm1)}{2}$ 
Defined solutions of Pell's equation:  $p^2-2s^2=\pm1$ 
But it is necessary to write the formula describing their solutions through solving Pell's equation: 
$X=p^2+4ps+4s^2$  
$Y=p^2+3ps+2s^2$ 
And more. 
$X=2s^2$ 
$Y=ps$ 
$p,s$ - These numbers can be any character. 
If you need to have a solution of the equation:  $Y^2=\frac{X(X\pm{a})}{2}$  
It is necessary to substitute into the formulas equation Pellya solutions:  $p^2-2s^2=\pm{a}$
