How can I find a nontrivial perfect square of a given quadratic form?

To translate your problem, you want to find an integer x such that

z^2 + 180z - 88 = x^2

Let's complete the square over the left hand side.

z^2 + 180z + 8100 - 88 = x^2 + 8100
(z + 90)^2 - 88 = x^2 + 8100

Therefore, moving the x^2 to the left control side, and the -88 to the becomingly control side,

(z + 90)^2 - x^2 = 8188

Factoring the left control side as a difference of squares, we obtain

([z + 90] - x) ([z + 90] + x) = 8188

(z + 90 - x) (z + 90 + x) = 8188

Now, we determine the coach factorization of 8188.

8188 = 2 x 4094 = 2 x 2 x 2047 = 2 x 2 x 23 x 89

Note such the couple (z + 90 - x) and (z + 90 + x) duty the couple be even. Out of the coach factors, the only bifid ways to do that is via the combination (2 x 23)(2 x 89) = (46)(176) and (2)(2x23x89) =
(2)(2 x 4094)

This parcel out us this only two accessible combinations:

(z + 90 - x) (z + 90 + x) = (46)(176)
(z + 90 - x) (z + 90 + x) = (2)(4094)

Now, we can equate each of them componentwise also we get a structure of equations.

(1) (z + 90 - x) (z + 90 + x) = (46)(176) means

z + 90 - x = 46
z + 90 + x = 176

(z = 21, x = 65)

(2) (z + 90 - x) (z + 90 + x) = (2)(4094) means

z + 90 - x = 2
z + 90 + x = 4094

(z = 1958, x = 2046)

Therefore, we inspire two values for z in order for z^2 + 180z - 88 to be a perfect square:

z = {21, 1958}

They create the conform precise squares 65^2 and 2046^2.

(EDIT: Giving credit to "a_math_guy",
(z + 90 - x) (z + 90 + x) = 8188

implies such both (z + 90 - x) and (z + 90 + x) duty be even.)