The general and probably complete solution to $a^2+(b-1)^2 = (b)^2 is (2v+1)^2+(2v(v+1))^2 = (v^2+(v+1)^2)^2$ We get the triples $(a,b-1,b) = (3,4,5), (5,12,13), (7,24,5^2),...,(41,840,29^2),...,(239,28560,13^4),...$
If we only want those b that are also squares, we must solve the Pell like equation $a^2 -2b^2 =-1$ whose $(a,b)$ solutions beginning with the trivial $(1,1)$ are $(7,5), (41,29), (239,13^2),...$ given by the recurrent relation :$ a(n+1) = 3a(n) + 4b(n); b(n+1) = 2a(n) + 3b(n)$. What is the general solution, if known, to $a^2 -2b^4 = -1 $?
Mordell, Diophantine Equations, page 271, writes, "...for the special case $$y^2=2x^4-1$$ it has been known for two centuries that solutions are given by $(x,y)=(1,1)$ and $(13,239)$. It was proved by Ljunggren that these are the only positive integer solutions. The proof is exceedingly complicated." Mordell gives the citation, Zur Theorie der Gleichung $x^2+1=Dy^4$, Avh. Norske Vid. Akad. Oslo No. 5 1 (1942).
Guy, Unsolved Problems In Number Theory, 3rd edition, Problem D6 is "An elementary solution of $x^2=2y^4-1$." Guy cites Steiner and Tzanakis, Simplifying the solution of Ljunggren's equation $X^2+1=2Y^4$, J Number Theory 37 (1991) 123-132, and writes, "Whether Steiner & Tzanakis have simplified the solution may be a matter of taste; they use the theory of linear forms in logarithms of algebraic numbers."
Guy also cites a proof by Chen Jian-Hua, which he calls "unconventional." The bibliographic details are, A new solution of the Diophantine equation $X^2+1=2Y^4$, J Number Theory 48 (1994) 62-74 and A note on the Diophantine equation $x^2+1=dy^4$, Abh. Math. Sem. Univ. Hamburg 64 (1994) 1-10.
I'd suggest also looking at Wikipedia on Ljunggren.
And there's more: Konstantinos A. Draziotis, The Ljunggren equation revisited, Colloq. Math. 109 (2007), no. 1, 9–11.
Michael A. Bennett, Irrationality via the hypergeometric method, Diophantine analysis and related fields—DARF 2007/2008, 7–18, AIP Conf. Proc., 976, Amer. Inst. Phys., Melville, NY, 2008.