The triangular-square problem, which marlu posted, leads to a concrete instance of Pell's equation. There are quite a few elementary problems leading to Pell's equation for small values of $d$: see Barbeau's book "Pell's Equation". These are nice problems, but because $d$ turns out to be a small explicit number the fact that a general $x^2 - dy^2 = 1$ with nonsquare $d > 1$ is nontrivially solvable is not directly relevant. Here is an example of a problem that involves infinitely many instances of Pell's equation, due to H. Alder and W. Simons ($n$ and $n+1$ Consecutive Integers with Equal Sums of Squares, Amer. Math. Monthly 74 (1967), 28--30.)
For a positive integer $n$ we would like to find a set of $n$ consecutive squares and $n+1$ consecutive squares whose sums are equal: $ x^2 + (x+1)^2 + \cdots + (x+n-1)^2 = y^2 + (y+1)^2 + \cdots + (y+n)^2 $ for some integers $x$ and $y$. (When $n = 1$ this is the problem of finding a Pythagorean triple with consecutive legs: 3,4,5; 20,21,29; 119,120,169;....)
Writing $z=x-y$, after some tedious algebra the above equation is the same as $ (y+n(1-z))^2 = n(n+1)z(z-1). $ Let $a^2$ be the largest square factor of $n(n+1)$: $n(n+1) = a^2b$, where $b$ is squarefree. Then $a^2b$ is a factor of $(y+n(1-z))^2$, so (!) $ab$ is a factor of $y+n(1-z)$. Set $y+n(1-z) = abv$, with $v$ an integer. Then the above equation is the same as $ a^2b^2v^2 = a^2bz(z-1) \Longleftrightarrow (2z-1)^2 - 4bv^2 = 1. $
We are thus reduced to solving $u^2 - 4bv^2 = 1$ for integers $u$ and $v$ (necessarily $u$ is odd), and $4b$ is not a square. Aside from the factor of 4, which could be absorbed into $v$ by insisting $v$ be even, this is a fairly general Pell equation with squarefree $b$. (One solution is $u=2n+1$ and $v = a$, leading to $x = 2n^2 + 2n+1$ and $y = 2n^2 + n$, but that is not always the first solution: try $n=8$.)
Moreover, every squarefree $b > 1$ does arise in such a problem, because there is always an integer $n$ such that $n(n+1) = a^2b$ for some integer $a$: finding $n$ and $a$ such that $n(n+1) = a^2b$ is equivalent to $(2n+1)^2 - 4ba^2 = 1$, and the equation $X^2 - 4bY^2 = 1$ does have some integral solution $(X,Y)$ by the general theory of Pell's equation, with $X$ necessarily odd.