Cesenatico

The notation 5654 in base $b$ represents the number $N=5 b^{3}+6 b^{2}+5 b+4=(b+1)\left(5 b^{2}+b+4\right)$. If $b$ is odd, $b+1$ is even; conversely, if $b$ is even, $5 b^{2}+b+4$ is even. In either case, $N$ is even, and therefore it is a power of 2.

It follows that $b+1$ and $5 b^{2}+b+4$ are both powers of $2$: $b+1=2^{n}$ and $5 b^{2}+b+4=2^{m}$. We note that $5 b^{2}+b+4 \geq b+4$ for every $b$, hence $m>n$. From the first equation we get $b=2^{n}-1$, and since $b>6$, then $n \geq 3$. Substituting into the second equation, we obtain: $5 \cdot 2^{2 n}-9 \cdot 2^{n}+8=2^{m}$. Dividing by 8 and rearranging, we get:
$$5 \cdot 2^{2 n-3}-2^{m-3}=9 \cdot 2^{n-3}-1 .$$
Since $m>n \geq 3$, the left-hand side of the equation is even, so $9 \cdot 2^{n-3}-1$ must also be even; it follows that $n=3$, and the equation becomes $5 \cdot 8-9+1=2^{m-3}$, from which $m=8$. Therefore the notation 5654 in base 7 represents $2048=2^{11}$, and $b=7$ is the unique solution of the problem.

The proof of divisibility by 11 for numbers written in decimal base can be mimicked as follows: $b \equiv-1(\bmod b+1)$, hence
$$N=5 b^{3}+6 b^{2}+5 b+4 \equiv 5(-1)^{3}+6(-1)^{2}+5(-1)+4 \equiv 0 \quad(\bmod b+1)$$
Since $(b+1) \mid N$, both $b+1$ and $N /(b+1)$ must be powers of the same prime. Now $N /(b+1)=5 b^{2}+b+4=(5 b-4)(b+1)+8$; it follows that the greatest common divisor between $b+1$ and $N /(b+1)$ must divide 8. But, between two powers of the same prime, the greatest common divisor is the smaller of the two numbers: in this case, evidently $b+1$. Therefore, since $b>6$, the only possibility is $b+1=8$, that is $b=7$ which, upon direct verification, gives a solution.