For suitable $a,b$, is it true that $(a+b) \nmid a^{m}$ for all $m \geq 2$?

Question

Let $a,b$ be non-coprime integers; let $a > b \geq 1$. Is it true that $$ (a+b) \nmid a^{m} $$ for all integers $m \geq 2$? This seems intuitively true. For instance, if $a := 4$ and if $b := 2$, then $a+b = 6$ does not divide $a^{m} = 16, 64, 256$ for $m = 2,3,4$ respectively.

Answer 1

If you use $a=10,b=6,m>4$, then $16\vert10^m$