I just wonder that by defining the positive integer $n$ that possesses the property $P$ to be the number that possesses $P$. For example, if $n$ possessing $P$ means that $n=x^2+y^2+b^k$ for fixed positive integer based $b$ and some nonnegative integers $x,y$. Furthermore let $n$ possessing $P'$ means the opposite that $n$ is not expressible by such form.
I think the work of Romanov on the number of the form $p+2^k$ has positive lower density almost implies that the number of the form $x^2+y^2+b^k$ for fixed based $b$ also has positive lower density. It is possible to show that the number not of the form has also positive lower density and that there is no arbitrary long consecutive numbers for which each of them not possesses $P'$. The proof is as follows.
We determine the case fixed based $m$. It can be shown that $x^2+y^2\pmod{m^2}$ for $0\le x,y\le m^2-1$ do not cover all residues modulo $m^2$ (for $m$ not being a product of primes of the form $4k+1$) and choose $m_1,m_2$ not being a prime of the form $4k+1$ where $\gcd(m,m_1)=\gcd(m,m_2)=\gcd(m_1,m_2)=1$. So we can choose the number of the form $a_0X+b_0$, for fixed $a_0,b_0$ and $X$ can be any natural number, which is not of the form $x^2+y^2+m^k$. We can take $a_0=(mm_1m_2)^2$ and $b_0$ by Chinese remainder theorem from the system to be the smallest (where $t,t_1,t_2 \not\equiv x^2+y^2$ for modulo $m,m_1,m_2$ respectively) $$b_0\equiv a_0X+b_0-m^k\equiv t\pmod{m^2}, k\ge 2,$$ $$b_0-m\equiv a_0X+b_0-m\equiv t_1\pmod{m_1^2}, k=1,$$ $$b_0-1\equiv a_0X+b_0-1\equiv t_2\pmod{m_2^2}, k=0.$$ Thus, for any $k$ we have $a_0X+b_0-m^k$ is not a sum of two squares. If $m$ is a product of primes of the form $4k+1$, we can choose $m-1=4k'$ (instead of $m^2$) so the number of residues posiible is $<4k'$ as there are $3$ out of $4$ numbers for modulo $4$ that is a sum of two squares and we can choose $t$ like before.
I wonder the other way around that, is it possible to find some fixed based $b$ so that there is arbitrary long consecutive numbers none of which is of the form $x^2+y^2+b^k$, i.e. Is there arbitrary long consecutive numbers for which each of them not posesses $P$ for some fixed $b$?