The first thing to note is that if $z$ is odd, then so is $x$ and vice versa because otherwise $z^2-x^2$ would be odd, and so not equal to $4y$ with $y\in\mathbb Z$. If both are even, then $z^2$ and $x^2$ are both divisible by 4 and so the equation has a unique solution. If both are odd, say $x=2n+1$ and $z=2m+1$, then $z^2-x^2=4(m^2-n^2+m-n)$ and the equation again has a unique solution.
Given some $x$ and the knowledge that $-N\leq y\leq N$, and letting $z\geq 0$, we can say though some real analysis that $z=\sqrt{x^2+4y}\leq x+\frac{2y}{x}$, so that $z\leq x+\frac{2N}{x}$. For the lower bound, if $x\leq \sqrt{N}$, then the argument of the square root can get less than $0$, so $z\geq 0$ is the best you can do. Otherwise, a simple linear bound is $\sqrt{x^2+4y}\geq x+\frac{4y}{x}$ for $0\leq y\leq x$ so that $z\geq x-\frac{4N}{x}$.
Computationally, you can use the simplicity of the solutions in $x$-$z$ space to enumerate the solutions: Given an $x$, start at $z=0$ or $1$ (or the other lower bound mentioned above) depending on whether $x$ is even or odd, solve for $y=\frac{1}{4}(z^2-x^2)$, and increase $z$ by twos until $y$ is greater than $N$. Doing this for $x=1$ to $N$ will give all possible solutions in the range you posed, although it may give extra solutions $y<-N$ if $x\geq\sqrt{N}$ as mentioned.