Fibonacci meets Triangular numbers

Consider the differences of the first two equations: $m+1 = {\color{#D61F06}t_{m+1}-t_{m}} = {\color{#3D99F6}f_{n+1}-f_{n}}={\color{#3D99F6}f_{n-1}}$

Consider the differences of the last two equations: $m+2 = {\color{#D61F06}t_{m+2}-t_{m+1}} = {\color{#3D99F6}f_{n+2}-f_{n+1}}={\color{#3D99F6}f_{n}}$

Consider the differences of the above two equations: $1 = {\color{#3D99F6}f_{n}-f_{n-1}}={\color{#3D99F6}f_{n-2}}$

Now ${\color{#3D99F6}f_{1}=f_2=1=f_{n-2}}$ implies that $n=3$ or $n=4$ .

If $n=3$ , the corresponding three consecutive Fibonacci numbers are $2, 3, 5$ , which the differences for consecutive 2 terms are 1 and 2. And the first few triangular numbers are $1,3,6,10,...$ which the differences for consecutive 2 terms are 2, 3, 4,.... It does not match.

Hence $n=4$ . It is easy to verify that $(m,n,k) = (1,4,2)$ is the only solution. $\begin{array} {llll} {\color{#D61F06}t_1}+2 &=&{\color{#D61F06}1}+2 &=& {\color{#3D99F6}3} & = & {\color{#3D99F6}f_4} \\ {\color{#D61F06}t_2}+2 &=&{\color{#D61F06}3}+2 &=& {\color{#3D99F6}5} & = & {\color{#3D99F6}f_5} \\{\color{#D61F06}t_3}+2 &=&{\color{#D61F06}6}+2 &=& {\color{#3D99F6}8} & = & {\color{#3D99F6}f_6}\\ \end{array}$

Watch this video for the explanation and discussion.