Solving diophantine equations -- does having a bound on the size of the solution help?
No.

If I follow you correctly, your argument is that if P=NP, then $\mathrm{RH}10\in \mathrm{P}$ and the same procedure would be a valid algorithm for $\mathrm{H}10$, which contradicts with the fact that $\mathrm{H}10$ is undecidable. 

The step that lacks proof is the claim that the implication:


  if $\mathrm{RH}10$ in P, then $\mathrm{H}10$ must necessarily be decidable


To see that this claim doesn't follow, consider what a polynomial algorithm to solve $\mathrm{RH}10$ could be doing. We only know it lies in P somehow, so let's assume it runs in $O(n^k)$ time for some $k\in\mathbb{N}$. Let $l$ be such that we can verify whether a value is a solution for a Diophantine equation with $n$ variables in $O(n^l)$ time.

It is possible that this is done by the following algorithm $A$, given an instance $I$ of $\mathrm{RH}10$:


Identify a set $S\subseteq [-2^n,2^n]^n$ of size $O(n^{k-l})$ in $O(n^k)$ time such that $s$ is a solution to the Diophantine equation encoded in $I$ if and only if $s\in S$.
Iterate over all elements in $s\in S$ and test whether it is a solution of $I$. If a solution is found, report that $I$ has a solution, otherwise report it has not.


Observe that this is a valid algorithm to solve $\mathrm{RH}10$ in $O(n^k)$ if and only if there exist a procedure to carry out step 1. We don't know whether this is true, it is unlikely, but we don't know so it is at least possible.

Now, does the existence of algorithm $A$ give us a procedure to decide $\mathrm{H}10$? No. Algorithm $A$ requires that the solution values are restricted to some finite subset of $\mathbb{Z}^n$ to work. This is true even if we only use step 1. I hope it is clear that the existence of algorithm $A$ brings us no closer to deciding $\mathrm{H}10$. 

Therefore, it is clear that the claim that 


  if $\mathrm{RH}10$ in P, then $\mathrm{H}10$ must necessarily be decidable


is not motivated, as the possibility of the existence of algorithm $A$ is not dis-proven, while we have seen that the existence of algorithm $A$ contradicts your claim.