We have $x=a^2$, and $2x+1=b^2$, so $b^2-2a^2=1$, and this is a Pell equation, with infinite many solution. On the other hand, we must have $m^2<2x+2$, and $3x+2<(m+1)^2$, so we get $\frac{{{m}^{2}}-2}{2}<{{a}^{2}}<\frac{{{(m+1)}^{2}}-2}{3}$ (*), so we have ${{m}^{2}}-4m-4<0$, thus $m\in \left\{ 0,1,2,3,4 \right\}$, and we must verify these numbers in (*), and we get the $a$, and the $x$.

$2\sqrt{2x+1}+1\geq x+1$ which holds for $\mathcal{O}(1)$ numbers.

Note that the next square number after $2x+1$ is $(\sqrt{2x+1} + 1)^2$. As none of the consecutive positive integers $2x+2, 2x+3, \ldots, 3x+2$ are squares, we must have $(\sqrt{2x+1} + 1)^2 > 3x+2$. This simplifies to $x^2-8x-4<0$. Hence, $(x-4)^2 <20$. As $x$ is natural, $\mid x-4\mid$ lies from $1$ to $4$. So, $x$ can take the values $1,2,3,4,5,6,7,8$. Out if these, only $4$ gives $2x+1$ to be a perfect square. Note that this solution for $x$ indeed satisfies the conditions. $\blacksquare$

TuZo wrote: We have $x=a^2$, and $2x+1=b^2$, so $b^2-2a^2=1$, and this is a Pell equation, with infinite many solution. On the other hand, we must have $m^2<2x+2$, and $3x+2<(m+1)^2$, so we get $\frac{{{m}^{2}}-2}{2}<{{a}^{2}}<\frac{{{(m+1)}^{2}}-2}{3}$ (*), so we have ${{m}^{2}}-4m-4<0$, thus $m\in \left\{ 0,1,2,3,4 \right\}$, and we must verify these numbers in (*), and we get the $a$, and the $x$. according to the problem statement, x doesn't have to be a perfect square, right?

Let $ 2x+1 = y^2$ so $y = \sqrt{2x+1}$. Note that whole problem says $ (y+1)^2 - y^2 > (3x+2) - (2x+1) = x+1$ so $ 2y+1 > x+1 $ so $2y > x$. $2\sqrt{2x+1} > x$ so $4(2x+1) > x^2$ so $4 > x(x-8)$ so $x < 9$. Now we can check that only for $ x = 4$ ,$2x+1$ is perfect square.