Is there a prime $m^2+n^2$ for all $m\in\mathbb N$ given $m \geq n$?

Question

It seems from a quick check that for all $m$, there is at least one $n$ where $m \geq n \geq 1$ and $m^2+n^2\in\mathbb P$.

Is this true, and is there a proof?

Empirically, it looks likely to be true, and the number of primes per given $m$ climbs fast. Furthermore, if it's provable that there are at least two solutions for any $m>12$ (which seems very likely heuristically), an alternative proof of Bertrand's postulate would immediately follow:

If we're guaranteed two prime pairs per $m$, the largest possible difference between two consecutive $m$ pairs changes from

$$m^2+1,\qquad(m+1)^2+m^2$$

to

$$m^2+4,\qquad(m+1)^2+(m-1)^2.$$

Then we have

$$2(m^2+4)>(m+1)^2+(m-1)^2=2(m^2+1),$$

and that should do it.

Answer 1

The number of solutions is given by OEIS/A069004. There it says that it is an open problem whether there are always solutions.