Below we only consider numbers $N=pq$, where $p$,$q$ are primes of the form $ (6j+1)$.
It's easy to show that $N + 9n^2 = d^2$. $9k^2$ is related to $(p-q)$ and $d^2$ is related to $(p+q)$. the $n$ in $9n^2$ is odd if $\dfrac{(N-1)}{6}$ is odd and even if $\dfrac{(N-1)}{6}$ is even. Below we consider the case of even $\dfrac{(N-1)}{6}$.
In this case we can rewrite $9n^2$ as $9(2n)^2=36n^2$. $d$ becomes $(6k+1)^2$ ( and $(6k-2)^2$ for the odd case ).
So we end up with the equation $N + 36n^2 = (6k+1)^2.$ $N$ can be written as $N=(6x+1)(6y+1)=36xy +6 (x+y) + 1$.
Let's consider $N=7 \cdot 199=1393=36 \cdot 38+25.$
$N + 36n^2 = (6k+1)^2$ becomes:
$36(n^2 + 38 ) + 24 +1 = 36k^2 + 12k + 1$
By equating coefficients, we can write: $n^2 + 38 = k^2 + l$ and $36l + 24 = 12k$ so that $3l + 2 = k.$ By substituting $k$ by its $l$ value in $n^2 + 38 = k^2 + l$ we get the final equation:
$n^2 = 9l^2 + 13l - 34$.
This polynomial in $l$ takes the value of $256=16^2$ for $l=5$.
Is there an efficient way to determine the $l$ value that makes the polynomial in $l$ a perfect square, that is without having to to check every single $l$ value?
Apparently you want to know when $9l^2 + 13 l - 34 $ is a square. The key point is that $9$ is a square. We can easily find inequalities that show the possible values for $l.$
Note $(3l+2)^2 = 9 l^2 + 12 l + 4.$ When is this less than $9l^2 + 13 l - 34 $ for positive $l?$ $$ 9 l^2 + 12 l + 4 < 9l^2 + 13 l - 34, $$ $$ 12 l + 4 < 13 l - 34, $$ $$ 38 < l.$$
Meanwhile, for positive $l,$ we always have $$ 9l^2 + 13 l - 34 < 9l^2 + 18 l + 9 = (3l+3)^2. $$ So, for $l \geq 39,$ we find $$ (3l+2)^2 < 9l^2 + 13 l - 34 < (3l+3)^2, $$ so the quantity in the middle is not a square because $3l+2, 3l+3$ are consecutive.
So we need only check $0 \leq l \leq 39.$ For $l \geq 3,$ we have $13 l > 34,$ and $9 l^2 + 13 l - 34 > (3l)^2.$ So, another approach is to check $l=0,1,2,$ then solve two equations for $l,$ $$ 9 l^2 + 13 l - 34 = (3l+1)^2 = 9 l^2 + 6l + 1, \; \; \; 13l-34 = 6l+1, \; \; 7l = 35, \; \; l=5, $$ $$ 9 l^2 + 13 l - 34 = (3l+2)^2 = 9 l^2 + 18l + 9, \; \; \; 13l-34 = 18l+9, \; \; -5l = 43, \; \; l= -43/5. $$
Similar work if you want to include negative $l$ on purpose.
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Note that, if the lead coefficient $9$ were to be replaced by something (positive) not a square, such as $2$ or $3,$ then either the polynomial is never a square or it is a square infinitely often. This is related to Pell type equations. For example, $2 l^2 + 1$ is a square infinitely often, because we are solving the Pell equation $n^2 - 2 l^2 = 1$ which has infinitely many solutions.