Solve in nonnegative integers the equation $5^{t}+3^{x} 4^{y}=z^{2}$.
If $x=0$ we have

$$
z^{2}-2^{2 y}=5^{t} \Longleftrightarrow\left(z+2^{y}\right)\left(z-2^{y}\right)=5^{t}
$$

Putting $z+2^{y}=5^{a}$ and $z-2^{y}=5^{b}$ with $a+b=t$ we get $5^{a}-5^{b}=2^{y+1}$. This gives us $b=0$ and now we have $5^{t}-1=2^{y+1}$. If $y \geq 2$ then consideration by modulo 8 gives $2 \mid t$. Putting $t=2 s$ we get $\left(5^{s}-1\right)\left(5^{s}+1\right)=2^{y+1}$. This means $5^{s}-1=2^{c}$ and $5^{s}+1=2^{d}$ with $c+d=y+1$. Subtracting we get $2=2^{d}-2^{c}$. Then we have $c=1, d=2$, but the equation $5^{s}-1=2$ has no solutions over nonnegative integers. Therefore so $y \geq 2$ in this case gives us no solutions. If $y=0$ we get again $5^{t}-1=2$ which again has no solutions in nonnegative integers. If $y=1$ we get $t=1$ and $z=3$ which gives us the solution $(t, x, y, z)=(1,0,1,3)$.

Now if $x \geq 1$ then by modulo 3 we have $2 \mid t$. Putting $t=2 s$ we get

$$
3^{x} 4^{y}=z^{2}-5^{2 s} \Longleftrightarrow 3^{x} 4^{y}=\left(z+5^{s}\right)\left(z-5^{s}\right)
$$

Now we have $z+5^{s}=3^{m} 2^{k}$ and $z-5^{s}=3^{n} 2^{l}$, with $k+l=2 y$ and $m+n=x \geq 1$. Subtracting we get

$$
2.5^{s}=3^{m} 2^{k}-3^{n} 2^{l}
$$

Here we get that $\min \{m, n\}=0$. We now have a couple of cases.

Case 1. $k=l=0$. Now we have $n=0$ and we get the equation $2.5^{s}=3^{m}-1$. From modulo 4 we get that $m$ is odd. If $s \geq 1$ we get modulo 5 that $4 \mid m$, a contradiction. So $s=0$ and we get $m=1$. This gives us $t=0, x=1, y=0, z=2$.

Case 2. $\min \{k, l\}=1$. Now we deal with two subcases:

Case 2a. $l>k=1$. We get $5^{s}=3^{m}-3^{n} 2^{l-1}$. Since $\min \{m, n\}=0$, we get that $n=0$. Now the equation becomes $5^{s}=3^{m}-2^{l-1}$. Note that $l-1=2 y-2$ is even. By modulo 3 we get that $s$ is odd and this means $s \geq 1$. Now by modulo 5 we get $3^{m} \equiv 2^{2 y-2} \equiv 1,-1(\bmod 5)$. Here we get that $m$ is even as well, so we write $m=2 q$. Now we get $5^{s}=\left(3^{q}-2^{y-1}\right)\left(3^{q}+2^{y-1}\right)$.

Therefore $3^{q}-2^{y-1}=5^{v}$ and $3^{q}+2^{y-1}=5^{u}$ with $u+v=s$. Then $2^{y}=5^{u}-5^{v}$, whence $v=0$ and we have $3^{q}-2^{y-1}=1$. Plugging in $y=1,2$ we get the solution $y=2, q=1$. This gives us $m=2, s=1, n=0, x=2, t=2$ and therefore $z=13$. Thus we have the solution $(t, x, y, z)=(2,2,2,13)$. If $y \geq 3$ we get modulo 4 that $q, q=2 r$. Then $\left(3^{r}-1\right)\left(3^{r}+1\right)=2^{y-1}$. Putting $3^{r}-1=2^{e}$ and $3^{r}+1=2^{f}$ with $e+f=y-1$ and subtracting these two and dividing by 2 we get $2^{f-1}-2^{e-1}=1$, whence $e=1, f=2$. Therefore $r=1, q=2, y=4$. Now since $2^{4}=5^{u}-1$ does not have a solution, it follows that there are no more solutions in this case.

Case $2 b . \quad k>l=1$. We now get $5^{s}=3^{m} 2^{k-1}-3^{n}$. By modulo 4 (which we can use since $0<k-1=2 y-2)$ we get $3^{n} \equiv-1(\bmod 4)$ and therefore $n$ is odd. Now since $\min \{m, n\}=0$ we get that $m=0,0+n=m+n=x \geq 1$. The equation becomes $5^{s}=2^{2 y-2}-3^{x}$. By modulo 3 we see that $s$ is even. We now put $s=2 g$ and obtain $\left(2^{y-1}-5^{g}\right)\left(5^{g}+2^{y-1}\right)=3^{x}$. Putting $2^{y-1}-5^{g}=3^{h}, 2^{y-1}+5^{g}=3^{i}$, where $i+h=x$, and subtracting the equations we get $3^{i}-3^{h}=2^{y}$. This gives us $h=0$ and now we are solving the equation $3^{x}+1=2^{y}$.

The solution $x=0, y=1$ gives $1-5^{g}=1$ without solution. If $x \geq 1$ then by modulo 3 we get that $y$ is even. Putting $y=2 y_{1}$ we obtain $3^{x}=\left(2^{y_{1}}-1\right)\left(2^{y_{1}}+1\right)$. Putting $2^{y_{1}}-1=3^{x_{1}}$ and $2^{y_{1}}+1=3^{x_{2}}$ and subtracting we get $3^{x_{2}}-3^{x_{1}}=2$. This equation gives us $x_{1}=0, x_{2}=1$. Then $y_{1}=1, x=1, y=2$ is the only solution to $3^{x}+1=2^{y}$ with $x \geq 1$. Now from $2-5^{g}=1$ we get $g=0$. This gives us $t=0$. Now this gives us the solution $1+3.16=49$ and $(t, x, y, z)=(0,1,2,7)$.

This completes all the cases and thus the solutions are $(t, x, y, z)=(1,0,1,3),(0,1,0,2)$, $(2,2,2,13)$, and $(0,1,2,7)$.

Note. The problem can be simplified by asking for solutions in positive integers (without significant loss in ideas).