= Solution
Put $\alpha=\sqrt N$. Every positive solution satisfies
$$
0<\frac xy-\alpha=\frac{M}{y^2(x/y+\alpha)}<\frac1{2y^2}.
$$
After reducing $x/y=p/q$, we still have $|\alpha-p/q|<1/(2q^2)$. Here is the <continued fraction convergent> criterion with its proof. If $p/q$ is reduced and obeys this inequality, it is a strict best approximation of the second kind: for any different $r/s$ with $0<s\le q$, the assumption $|s\alpha-r|\le|q\alpha-p|$ would give
$$
1\le|ps-qr|\le s|p-q\alpha|+q|r-s\alpha|<\frac{s+q}{2q}\le1,
$$
a contradiction. Choose consecutive <continued fraction convergents> $p_j/q_j,p_{j+1}/q_{j+1}$ with $q_j\le q<q_{j+1}$. Their <determinant> is $\pm1$, their errors $e_j=q_j\alpha-p_j$ alternate in sign, and every <integer> pair $(r,s)$ is uniquely $a(p_j,q_j)+b(p_{j+1},q_{j+1})$. For $0<s<q_{j+1}$, either $b=0$, or $a,b$ have opposite signs: positive $a,b$ would make $s\ge q_{j+1}$, negative $a,b$ make $s<0$, and $a=0$ is impossible. Opposite signs make $ae_j$ and $be_{j+1}$ have the same sign, so $|s\alpha-r|\ge|e_j|$. If $b=0$, reduction gives the $j$th convergent itself. Thus a strict best approximation must be a convergent. The <determinant> and alternating-error facts follow immediately from the recurrence $p_j=a_jp_{j-1}+p_{j-2}$, $q_j=a_jq_{j-1}+q_{j-2}$ and the positive continued-fraction tail formula. Therefore \b[the reduced ratio $x/y$ is a convergent of $\sqrt N$]. Without assuming $\gcd(x,y)=1$, the unreduced pair may be a common multiple of the convergent's numerator and denominator; for example $(10,2)$ for $N=24,M=4$ reduces to $5/1$.
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