A difficult Diophantine equation

This is not a complete answer, just a few observations way too long for a comment.

Let $a>1$ be an integer, and let $x$, $y$ and $z$ be integers such that $$a^x+(a+1)^y=(2a+1)^z.$$

Observation 1: $z>0$.

Proof. Suppose toward a contradiction that $z\leq0$. Then $(2a+1)^z\leq1$, and because $a^x$ and $(a+1)^y$ are positive it follows that $a^x<1$ and $(a+1)^y<1$, meaning that $x,y<0$. Then clearing denominators yields the polynomial equation $$(a+1)^{-y}(2a+1)^{-z}+a^{-x}(2a+1)^{-z}=a^{-x}(a+1)^{-y},$$ where $-x,-y>0$ and $-z\geq0$. We can factor the left hand side to get $$(2a+1)^{-z}(a^{-x}+(a+1)^{-y})=a^{-x}(a+1)^{-y}.$$ The two factors on the right hand side are both coprime to $(2a+1)^{-z}$, so this forces $(2a+1)^{-z}=1$ and hence $z=0$. Then we are left with $$a^{-x}+(a+1)^{-y}=a^{-x}(a+1)^{-y},$$ but the unique solution to $u+v=uv$ in positive integers is is $u=v=2$, which shows that the equation above has no integral solutions, a contradiction. We conclude that $z>0$.$\qquad\square$

Observation 2: $x,y>0$ unless $(a,x,y,z)=(2,2,0,1)$.

Proof. From $z>0$ it follows that $(2a+1)^z>2$. If $x\leq0$ and $y\leq0$ then $$(2a+1)^z=a^x+(a+1)^y\leq1+1=2,$$ a contradiction. So either $x>0$ or $y>0$. If $y>0$ then $x\geq0$ because $$a^x=(2a+1)^z-(a+1)^y,$$ is an integer. Reducing mod $a$ shows that $x>0$.

Similarly, if $x>0$ then $y\geq0$ for the same reason as before. If $y=0$ then $$(2a+1)^z=a^x+(a+1)^y=a^x+1,$$ which shows that $x>1$ as $(2a+1)^z>a+1$, and that $a$ is even. By the binomial theorem we get $$1=(a+1)^y=-a^x+\sum_{i=0}^z\binom{z}{i}(2a)^i,$$ and hence rearranging the terms and dividing by $a$ shows that $$a^{x-1}=\sum_{i=1}^z\binom{z}{i}(2a)^{i-1}.$$ Because $x>1$ the left hand side is divisible by $a$, and clearly all terms on the right hand side with $i>1$ are divisible by $a$, hence so is the term with $i=1$, which is $\binom{z}{1}=z$. In particular $z$ is also even, say $z=2w$, so $$a^x=(2a+1)^z-1=((2a+1)^w-1)((2a+1)^w+1).$$ Of course the gcd of the two factors on the right hand side divides $2$, and because the two factors are congruent mod $2$ and the left hand side is even, their gcd is precisely $2$. Then by unique factorization we have $$(2a+1)^w-1=2^mu^x\qquad\text{ and }\qquad (2a+1)^w+1=2^nv^x,$$ for odd coprime positive integers $u$ and $v$, and positive integers $m$ and $n$ with $m=1$ or $n=1$. It follows that $$1=\frac{((2a+1)^w+1)-((2a+1)^w-1)}{2}=\frac{2^nv^x-2^mu^x}{2}=2^{n-1}v^x-2^{m-1}u^x,$$ which shows that $m=n=1$ and $x=2$. We are left with $$a^2+1=(2a+1)^z,$$ which forces $z<2$ so $z=1$, from which we find $a=2$. This shows that the only solution $(a,x,y,z)$ that does not satisfy $x,y>0$ is $(2,2,0,1)$.$\qquad\square$

A quicker way to 'show' that this is the only solution with $xy=0$ is to apply Mihăilescu's theorem: If either $x=0$ or $y=0$ then $$(2a+1)^z-(a+1)^y=1\qquad\text{ or }\qquad (2a+1)^z-a^x=1,$$ respectively, and because $2a+1>3$ this implies that either $z<2$ or $x,y<2$, from which the proof is quickly finished. But this result seems too advanced to use for an exercise like this.

Observation 3: If $a=2$ then either $x=y=z=1$ or $x$, $y$ and $z$ are all even.

Proof. If $a=2$ and $x\geq3$ then reducing mod $8$ and mod $3$ the equation $$2^x=5^z-3^y,$$ shows first that $y$ and $z$ are both even, and then that also $x$ is even.

If $x=2$ then $4=5^z-3^y$ and reducing mod $8$ shows that $z$ is odd and $y$ is even, say $y=2v$. Then $$5^z=4+3^y=(2+3^vi)(2-3^vi),$$ and the unique factorization $5=(2+i)(2-i)$ in $\Bbb{Z}[i]$ shows that $v=0$, contradiction observation $2$.

If $x=1$ then $2=5^z-3^y$ and reducing mod $8$ shows that $y$ and $z$ are both odd, and reducing mod $7$ shows that $y\equiv z\equiv 1\pmod{6}$. Then $y=1$ as otherwise $5^z\equiv2\pmod{9}$, which implies $z\equiv5\pmod{6}$, a contradiction. It is then clear that also $z=1$.$\quad\square$

Observation 4: $x\equiv z\pmod{2}$.

Proof. Reducing mod $a+1$ shows that $$(-1)^x+0^y\equiv(-1)^z\pmod{a+1},$$ and $a+1>2$, so the result follows.

Observation 5: If $z>1$ then $x>z$ or $y>z$ and $x,y<2z$.

Proof. If $x,y\leq z$ and $z>1$ then by the binomial theorem $$(2a+1)^z> a^z+(a+1)^z\geq a^x+(a+1)^y.$$ If $a>2$ and $x\geq2z$ then $a^2>2a+1$ and so $$a^x\geq(a^2)^z>(2a+1)^z,$$ and similarly $y\geq2z$ leads to a contradiction.