This seems a perfectly reasonable question to ask for solutions in integers. I cannot find any mention of it - not to say somebody has not considered it.

Consider first  x² + y²/N = b².

Let z = y / x  and  w = b / x ,  giving 1 + z² / N = w².  This has the obvious solution z=0, w=1.

The line of gradient m through this point meets the quadratic again at the point
z = 2 m N / ( 1 - m² N ).

Let m = p / q, and using  z = y / x, this gives the parameterisation of  x² + y² / N = b² as
x = q² - N p²   and  y = 2 N p q.

Substitute this into the equation  x² + N y² = a², and we get

a2  =  N2  p4  +  ( 4 N3 - 2 N ) p2 q 2  + q4

and define   g = N a / q²  and  h = N p / q giving

g2 = h4 + 2 N ( 2 N2 - 1 ) h2 + N2

This quartic has the clear rational solution  h = 0 , g = N , and so is birationally equivalent to an elliptic curve.

We find the family of curves

EN :   i2  =   j3  +  N ( 1 - 2 N2 ) j2   +  N4 ( N 2 - 1 ) j

with the transformation   h = N p / q = i / j.

The RHS of E_N factors as   j ( j - N³ ) ( j - N ( N² - 1 ) ) so there are 3 finite torsion points of order 2.

If N is a square then there are also 4 points of order 4, but I am unable to prove these are the only torsion points. None of these points lead to a non-trivial solution to the problem so we need E_N to have rank at least 1.

As an example,  N = 19 has rank 1 with a point   j = 95 * 17803² / 2093²
which gives  p = 27302 and q = 1961141, leading to
x = 3831911437005  and  y = 2034636720116.

We used the Birch and swinnerton-Dyer conjecture to to predict the rank of the curves for N in the interval  [ 2 , 249 ]. The curves with rank > 0 give the results in the file dd8res .

There are currently 8 values of N where we think that the curve has rank 1, but we are unable to find a generator of infinite order. The relevant values are given in the following table: