Maximum number of positive integer solutions.

The equation $\frac{2^n+b^n}{2^{n-1}+b^{n-1}}=m\quad\quad\quad\quad (*)$ can be transformed into $(\frac{b}{2})^{n-1}=\frac{m-2}{b-m},$ and let us assume that $n$ and $m$ are both positive integers. Since the left side of the previous equation is always positive, then the right side has to be also positive and, therefore, $2<m<b.$ A consequence of this is that if $b=3,$ there is not a possible integer value of $m$ between 2 and 3, and this would imply that the given equation does not have any solution $(n, m)$ in the case that $b=3.$ Therefore, we can assume that $b \geq 4.$

Additionally, since $m\leq b-1,$ then $\frac{m-2}{b-m}\leq \frac{b-1-2}{b-(b-1)}\leq b-3.$ Thus $(\frac{b}{2})^{n-1}\leq b-3.$ Taking natural log of both sides of the equation and solving for $n,$ we obtain that $n\leq \frac{\ln(b-3)}{\ln b-\ln 2}+1 <\frac{\ln(b)}{\ln b-\ln 2}+1=2+\frac{\ln 2}{\ln b -\ln 2}.$ Since $\frac{\ln 2}{\ln b -\ln 2}\leq 1,$ whenever $b\geq 4,$ and $n$ is a positive integer, then $n\leq 2.$ Then the solutions of the equation (*) must correspond to $n=1$ or $n=2$

Notice that when $n=1,$ the equation $(*)$ has a solution $(1, 1+b/2),$ if and only if $b$ is an even number. Let us assume that $n=2.$ Then the equation (*) becomes, $\frac{2^2+b^2}{2+b}=m$ where $m$ is an integer number. From the previous equation, we get that $\frac{8}{b+2}=m-b+2.$ Therefore , $b+2$ must be a factor of 8, and since $b\geq 4,$ the only possible value of $b$ is 6, and in this case the given equation would have the solution $(2, 5).$

Therefore, the equation (*) has two solutions only in the case that $b=6.$ Otherwise, it has one solution or no solution. Therefore, the answer for this question is $\boxed{2}.$

Let 2^n+b^n=An

Now if An-1 divides An then ; An-1|An;

=>An-1|An-1 A1-(b 2^(n-1)+2*b^(n-1)

=>An-1 |(b 2^(n-1)+2 b^(n-1)

=>An-1 k=(b 2^(n-1)+2*b^(n-1)

=>(K-2)b^(n-1)-b 2^(n-1)+k 2^(n-1)=0

=>2^(n-1)>b^(n-2), for k>2 as for k=1 or k=2 ,b=2/3 and 2 respectively.

So one possible case of b is b=3; But this doesn't have any solution for m when n= 1,2,3.

So the other case would be n=2 ,b=6 , Also n=1 ,b=6. So this is the only possible case with b>2 and it has 2 different sets of (n,m).