General formula to obtain triangular-square numbers I am trying to find a general formula for triangular square numbers. I have calculated some terms of the triangular-square sequence ($TS_n$):
$TS_n=$1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056
1882672131025, 63955431761796, 2172602007770041, 73804512832419600
I have managed to find that:
$m^2 = \frac{n(n+1)}{2}$
where $m$ is the $m^{th}$ term of the square number sequence, and $n$ is the $n^{th}$ term of the triangular numbers sequence.
Would anyone be able to point out anything that could help me derive a formula for the square-triangular sequence?

$${t_n} = \sum\limits_{k = 1}^n k  = 1 + 2 + 3 +  \cdots  + n = \frac{{n\left( {n + 1} \right)}}{2} = n-th{\text{ triangular number}}$$
$${s_m} = {\text{m-th square number}} = {m^2}$$
$${s_m} = {t_n} \Rightarrow \frac{1}{2}n\left( {n + 1} \right) = {m^2}$$
$$\frac{1}{2}{\left( {n + \frac{1}{2}} \right)^2} = \frac{1}{2}\left( {{n^2} + n + \frac{1}{4}} \right) = \frac{1}{2}\left( {{n^2} + n} \right) + \frac{1}{8}$$
$$\frac{1}{2}n\left( {n + 1} \right) = \frac{1}{2}\left( {{n^2} + n} \right) = \frac{1}{2}{\left( {n + \frac{1}{2}} \right)^2} - \frac{1}{8} = {m^2}$$
$$\frac{1}{2}{\left( {n + \frac{1}{2}} \right)^2} - {m^2} = \frac{1}{8}$$
$$4{\left( {n + \frac{1}{2}} \right)^2} - 8{m^2} = 1$$
$$2 \cdot \left( {n + \frac{1}{2}} \right) \cdot 2\left( {n + \frac{1}{2}} \right) - 8{m^2} = 1$$
$${\left( {2n + 1} \right)^2} - 8{m^2} = 1$$
$${\left( {2n + 1} \right)^2} - 2 \cdot {\left( {2m} \right)^2} = 1$$
$$\boxed{w \equiv 2n + 1}$$
$$\boxed{z \equiv 2m}$$
$$\boxed{{w^2} - 2{z^2} = 1}$$
Finding numbers that satisfy the last equation above isn't all that simple... so I found the first numbers that are both triangular and square using python
import pandas as pd
import numpy as np
triangluar_and_square = []
for n in np.arange(1,10000):
    w = 2*n + 1
    for m in np.arange(1,10000):
        z = 2*m
        if w*w - 2*z*z - 1 == 0:
            triangluar_and_square.append([m*m,m,n])
triangular_and_square = pd.DataFrame(triangluar_and_square,
                                     columns = ["Square (m-squared) + Trianglular","m","n"], 
                                     index = [1,2,3,4,5,6])

print(triangular_and_square)`

Code output
$$\begin{array}{*{20}{c}}
& {{\rm{Square (}}{m^2}){\rm{ and Triangular}}}&& m&& n& \\
\hline
& 1&& 1&& 1& \\
& {36}&& 6&& 8& \\
& {1225}&& {35}&& {49}& \\
& {41616}&& {204}&& {288}& \\
& {1413721}&& {1189}&& {1681}& \\
& {48024900}&& {6930}&& {9800}& 
\end{array}$$
