$1; 841; 1681$ is a 3-term AP of squares ($1^2;29^2,41^2$).
Add $528$ to each to get $529; 1369; 2209$, another 3-term AP of squares ($23^2;37^2;47^2$).
Some idea of how I found this:
First, I looked for 3-term APs of squares of the form $1<a^2<b^2$. This gives $2a^2-b^2=1$, essentially a Pellian, with solutions related to the convergents to the continued fraction for $\sqrt2$. The convergent $7/5$ leads to $1^2,5^2,7^2$; the next solution comes from the convergent $41/29$, $1^2,29^2,41^2$.
This AP has common difference $840$, so I looked for another 3-term AP of squares, $r^2<s^2<t^2$, with this common difference. So we need $s^2-r^2=t^2-s^2=840$. Said another way, we need two expressions for $840$ as a difference of two squares, such that the bigger square in one expression is the smaller square in the other.
So, we make a list of all the ways to write $840$ as a difference of two squares. There is a standard technique for this. Factor $840=8\times3\times5\times7$, and write down all the ways to write $840=uv$ with $u$ and $v$ both even, $u\le v$; for each such factorization, we get $840=((v+u)/2)^2-((v-u)/2)^2$.
$840=10\times84$ yields $840=47^2-37^2$, and $840=14\times60$ yields $840=37^2-23^2$, and Bob's your uncle.
Additional information, 24 January 2025:
There's more to say about three-term arithmetic progressions of squares, but first a review of Pythagorean triples, which turn out to be closely related to, but better studied than, three-term arithmetic progressions of squares.
A Pythagorean triple is a triple $(a,b,c)$ with $0<a<c$ and $0<b<c$ and $a^2+b^2=c^2$; a triangle with sides $a,b,c$ will be a right triangle with hypotenuse $c$. If $0<n<m$, then $(2mn,m^2-n^2,m^2+n^2)$ is a Pythagorean triple. The area of the corresponding right triangle is $(ab)/2$, which is $mn(m^2-n^2)$.
If $m,n$ are integers, relatively prime, and of different parity, then the corresponding Pythagorean triple is primitive, meaning the three components are relatively prime.
The relation to three-term arithmetic progressions of squares is that if $(a,b,c)$ is a Pythagorean triple, then $(|b-a|,c,b+a)$ is a triple of numbers whose squares are in AP. For example, from the Pythagorean triple $(3,4,5)$, we get the triple $(1,5,7)$, whose squares ($1$, $25$, and $49$) are in AP.
In terms of $m,n$, we get the triple $(|m^2-2mn-n^2|,m^2+n^2,m^2+2mn-n^2)$, whose squares are in AP.
The common difference of the three-term AP is $4mn(m^2-n^2)$, exactly four times the area of the corresponding right triangle.
Now, we've seen that to get two three-term APs of squares such that there is a single number that we can add to each of the three terms in one AP to get the three terms of the other AP, it is necessary and sufficient that the two APs have the same common difference. In terms of triangles, we are looking for two right triangles with the same area.
Time for an example. $(m,n)=(10,3)$ generates the right triangle $(60,91,109)$; $(m,n)=(14,1)$ generates the right triangle $(28,195,197)$. Both triangles have area $mn(m^2-n^2)=2730$.
The first triangle relates to the triple $(31,109,151)$ whose squares, $31^2=961$, $109^2=11881$, and $151^2=22801$, are in AP with common difference $4\times2730=10920$. The second triangle relates to the triple $(167,197,223)$, whose squares, $167^2=27889$, $197^2=38809$, and $223^2=49729$, are in AP, also with common difference $10920$. And if you add the number $26928$ to each of the first three squares, you get the second three squares, giving us another example of the relation requested in the original question.
Now, we could have done all those calculations without any mention of Pythagorean triples and right triangles. But here's the point: so far as I know, no one has ever looked at the question OP raises here, but many people have looked at the question of equiareal right triangles! Since we've shown that the two questions are equivalent, we can use their work to find solutions to our problem.
In particular, there is some discussion of equiareal right triangles in Beiler, Recreations in the Theory of Numbers, pages 126-7. For example, we are told that the smallest area common to three primitive Pythagorean triangles is $13123110$. The three Pythagorean triples are $(4485,5852,7373)$, generated by $(77,38)$; $(19019,1380,19069)$, by $(138,5)$; and $(3059,8580,9109)$, by $(78,55)$. So we get three triples of numbers whose squares are in AP, with the same common difference; $(1367,7373,11858)$, $(17639,19069,20399)$, and $(5521,9109,11639)$. And these give us three numbers $r,s,t$ and two numbers $u,v$ such that $(r,s,t)$, $(r+u,s+u,t+u)$, and $(r+v,s+v,t+v)$ are all three-term APs of squares, where OP was only asking for one such number $u$.
Beiler also gives an example of four equiareal right triangles, which lead to four three-term APs of squares with the same common difference. I'll just give the $m,n$ values of the triangles: $56,55$; $40,37$; $37,7$; $37,33$. The common area is $341880$.
Beiler gives a list of $32$ references at the end of the ($31$ page) chapter, but doesn't indicate which reference contains which result, so I don't know where to find the originals. My best guess is that they'll be in Ginsburg, Triplets of equiareal rational triangles, Scripta Math. 20 (1954) 219. But good luck finding issues of Scripta Math.
