More generally, I assert:
$$\pi(na^k)-\pi(na^{k-1})\leq \pi(na^{k+1})-\pi(na^k)\quad\text{for }n,a,k \in \mathbb N.$$
The most useful case of this is probably to say that if there are $m$ primes in $[n,2n]$, then there will be at least $m$ primes in $[2n,4n]$. I thought it worth noting this can also be written as
$$\pi(2n) \leq \frac{\pi(4n)+\pi(n)}{2}.$$
I doubt there are counterexamples, but if someone finds one, please share. Barring that, since this seems like an intermediate-strength statement, I'm curious how easy or difficult this might be to prove, and whether there are known similar results.
Update
Looks like a much tighter bound can also be specified below as $ \pi(3n)+\pi(n) \leq 2\pi(2n)$ for $n>1323$.
Also, note this can be used in reverse.
E.g. suppose we've worked out that the only primes in $[8,16]$ are $11$ and $13$. Then, even knowing nothing whatsoever about primes other than this rule (and that they start at $2$), it follows that there could be at most $2$ more primes in $[4,8]$, and another $2$ in $[2,4]$, so we'd know that $\pi(13)\leq 6$.
One readily tests that $$ 2\pi(2n)\le\pi(4n)+\pi(n)$$ holds at least for $1\le n<60184$. Following [Pierre Dusart, 2010], we know that $$ \frac x{\ln x-1}<\pi(x)<\frac x{\ln x -1.1}$$ for all $x\ge60184$. Hence at least for $n\ge 60184$, we have $$ \pi(2n)<\frac{2n}{\ln 2n -1.1}$$ and $$ \pi(4n)+\pi(n)>\frac{4n}{\ln 4n-1}+\frac n{\ln n-1}.$$
For the claim it would be sufficient to have $$ \frac{4n}{\ln 2n-1.1}<\frac{4n}{\ln 4n-1}+\frac n{\ln n-1}.$$ If we multiply by $\frac 1{4n}$ and substitute $\ln n=t+1$, our goal becomes $$ \frac{1}{t +\ln 2-0.1}<{\frac{1}{t+\ln 4}+\frac 1{4t}}$$ for all $t\ge \ln 60184-1\approx 10$. But this follows directly from $$ \frac{1}{t +\ln 2-0.1}-\frac{1}{t+\ln 4} =\frac{\ln 2 +0.1}{(t +\ln 2-0.1)(t+\ln 4)}<\frac1{t^2}<\frac1{4t}.$$
Replacing $2$ with an integer $a\ge2$, the same method boils down to $$ \frac{2a}{\ln a+t-0.1}-\frac{a^2}{2\ln a+t}<\frac {1}{t}$$ which is readily shown. However, the "base" case for $n<60184$ but with arbitrary $a$ is not so trivial.