Are there finitely many Pythagorean triples whose smallest two numbers differ by 1?

There is an infinite number of such pythagorean triples. Any primitive pythagorean triple $(a,b,c)$ with $a^2+b^2=c^2$ (are we are looking for primitive triples since we want $a$ and $b$ consecutive) is of the form: $$ a=p^2-q^2,\qquad b=2pq,\qquad c=p^2+q^2 $$ with $p$ and $q$ coprime and not both odd. So we are looking for integer solutions of: $$ p^2-2pq-q^2 =\pm 1,$$ or: $$ (p-q)^2 - 2q^2 = \pm 1.$$ However we know that the Pell equation $A^2-nB^2=1$ has an infinite number of integer solutions $(A,B)$ for every $n$ that is not a square, hence we can find "consecutive" pythagorean triples from the solutions of $$ A^2 - 2B^2 = 1,\tag{1}$$ for istance. $(A,B)=(3,2)$ is the minimal solution of $(1)$, giving $(p,q)=(5,2)$, hence the triple $(20,21,29)$. The next solution can be found by expanding: $$ (3+2\sqrt{2})^2 = 17+12\sqrt{12},$$ hence $(p,q)=(29,12)$ gives the triple $(696,697,985)$ and so on.

In general, we can see that all the solutions depends on the convergents of the continued fraction of $\sqrt{2}$, i.e. on the Pell sequence: $$ (p,q) = (P_n,P_{n+1}),$$ from which:

$$ (a_n,b_n) = (2P_nP_{n+1},P_{n+1}^2-P_{n}^2) = (2P_n P_{n+1},2P_nP_{n+1}+(-1)^n),$$

where:

$$P_n = \frac{1}{2\sqrt{2}}\left((1+\sqrt{2})^n-(1-\sqrt{2})^n\right).$$

Since neither $2P_n P_{n+1}$ or $P_{n+1}^2-P_{n}^2$ may be primes if $n> 1$, the only "consecutive" pythagorean triple with the smallest element being a prime is $(3,4,5)$.

Let's suppose $(x,x+1,y)$ is a Pythagorean triple. Then:

$x^2+(x+1)^2 = y^2$

$2x^2+2x+1 = y^2$

$4x^2+4x+2 = 2y^2$

$(2x+1)^2 + 1 = 2y^2$

$(2x+1)^2 - 2y^2 = -1$

This is now a negative Pell's Equation for which there are infinitely many solutions. See this question for details.

Cheap version: we get triples $(x,y,z)$ such that $y=x+1$ and $x^2 + y^2 = z^2,$ with $$ (3 ,4,5 ) $$ $$ (20 ,21,29 ) $$ $$ ( 119,120, 169 ) $$ $$ (696 ,697, 985 ) $$ $$ (4059 , 4060,5741 ) $$ $$ (23660 ,23661,33461 ) $$ $$ (137903 ,137904,195025 ) $$

That is, $(x_n, y_n,z_n)$ such that $$ x_{n+2} = 6 x_{n+1} - x_n +2, $$ but $$ y_{n+2} = 6 y_{n+1} - y_n -2, $$ and $$ z_{n+2} = 6 z_{n+1} - z_n . $$

If you put them the way I had them at first, odd first and even second, you get a $4 (-1)^n$ addend that I thought was a bit too much work.