Prime basis

In polynomial notation $(5654)_b=5\times b^3+6\times b^2+5\times b+4=(b+1)(5b^2+b+4)$

Let $(5654)_b=n^k$ where n is a prime number

$\implies b+1=n^i,5b^2+b+4=n^j$ where $i+j=k$

Also $\frac{5b^2+b+4}{b+1}=n^{j-i}$ $\implies$ remainder=0.

But by remainder theorem the above equation gives a remainder 8 ( put $b=-1$ ). To satisfy the above condition (remainder should be 0 ), 8 should be a multiple of $b+1$ and $b>6$ $\implies \boxed{b+1=8}$