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We use the following fact without proof. [Fact] (You can prove it by yourself) $a_1=1, a_2=7, b_1=1, b_2=5$ $a_{m+2}=6a_{m+1}-a_m$ $b_{m+2}=6b_{m+1}-b_m$ $(a,b)=(a_m,b_m)$ is a positve solution of the equation. From the above the fact, we can get $c_{m+2}=6c_{m+1}-c_m$ where $c_m=a_m-b_m$. Note that coefficients of $c_{m+2}$ and $c_m$ are $1$ or $-1$. So we can say the sequence $(c_m)$ has perfect period of modulo $n$. Since $c_1=a_1-b_1=0$, we can find positive integer $n>1$ such that $c_m \equiv 0$ (mod $n$)

prove that there are infinitely (a,b) and use principle pigeonhole. pd: Peru TST

@above that is not necessarily sufficient. For example, what if none of the infinite pairs of $(a,b)$ satisfied $a\equiv b\mod n$. @2above could you elaborate on what you mean by perfect period and why that it is true?

This is a shortest solution: Just take $a=2nk+c$ and $b=nk+c$