First of all, the term "pairs" is two of them, I assume (question's formulation is rather difficult to understand for me). So I guess this is the statement: $$\text{There exist infinitely many pairs of consecutive squares }x^2,\ (x+1)^2\text{ s.t. }x^2+(x+1)^2=y^2\ x,y\in\mathbb{N}.$$
So there are two theorems I know of that might be of any use here:
1. An odd prime $p$ can be written as sum of squares iff $p\equiv1$ mod $4$.
2. $n\in\mathbb{N}$ is the sum of squares iff primes that are $3$ mod $4$ occur an even number of times in the prime factorisation of $n$.
We can work out the LHS like so: $x^2+(x+1)^2=2x(x+1)+1$. We know that either $x$ or $x+1$ must be even; hence, the expression is of the form $4k+1$ with $k\in\mathbb{Z}$. From this we can conclude that the sum of two consecutive squares is always congruent $1$ modulo $4$ which also implies that primes that are $p\equiv3$ mod $4$ that divide the sum of the consecutive squares must occur an even number of times. To this point, this does not prove anything significant, I think.
Does anyone have a hint on how to prove the statement?
This question was answered here, and in Galc127's comment above. Basically, the above relation is the pell equation $2y^2-(2x+1)^2=1$, and the cool thing about pell equations is that when you have one solution, you have a countable infinity of them, visibly demonstrated by the identity below:
$$2(y)^2-[2x+1]^2=1=2(3y+4x+2)^2-[4y+6x+3]^2=1 \\ \implies \text{The recursive relation:} $$
$$\begin{cases} y_{k+1}=3y_k+4x_k+2 \\ x_{k+1}=2y_k+3x_k+1 \end{cases}$$